Answer:
1.07 × 10⁸ m/s
Explanation:
Using the relativistic Doppler shift formula which can be expressed as:
[tex]\lambda_o = \lambda_s \sqrt{\dfrac{c+v}{c-v}}[/tex]
here;
[tex]\lambda _o[/tex] = wavelength measured in relative motion with regard to the source at velocity v
[tex]\lambda_s =[/tex] observed wavelength from the source's frame.
Given that:
[tex]\lambda _o[/tex] = 656.3 nm
[tex]\lambda_s =[/tex] 953.3 nm
We will realize that [tex]\lambda _o[/tex] > [tex]\lambda_s[/tex]; thus, v < 0 for this to be true.
From the above equation, let's make (v/c) the subject of the formula: we have:
[tex]\dfrac{\lambda_o}{\lambda_s}=\sqrt{\dfrac{c+v}{c-v}}[/tex]
[tex]\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2=\dfrac{c+v}{c-v}[/tex]
[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2-1}{\Big(\dfrac{\lambda_o}{\lambda_s} \Big)^2+1}[/tex]
[tex]\dfrac{v}{c} =\dfrac{\Big(\dfrac{656.3}{953.3} \Big)^2-1}{\Big(\dfrac{656.3}{953.3} \Big)^2+1}[/tex]
[tex]\dfrac{v}{c} =0.357[/tex]
v = 0.357 c
To m/s:
1c = 299792458 m/s
∴
0.357c = (299 792 458 × 0.357) m/s
= 107025907.5 m/s
= 1.07 × 10⁸ m/s
A laser of wavelength 720 nm illuminates a double slit where the separation between the slits is 0.22 mm. Fringes are seen on a screen 0.85 m away from the slits. How far apart are the second and third bright fringes
Answer:
The appropriate solution is "2.78 mm".
Explanation:
Given:
[tex]\lambda = 720 \ nm[/tex]
or,
[tex]= 720\times 10^{-9} \ m[/tex]
[tex]D=0.85 \ m[/tex]
[tex]d = 0.22 \ mm[/tex]
or,
[tex]=0.22 \times 10^{-3} \ m[/tex]
As we know,
Fringe width is:
⇒ [tex]\beta=\frac{\lambda D}{d}[/tex]
hence,
Separation between second and third bright fringes will be:
⇒ [tex]\theta=\beta=\frac{\lambda D}{d}[/tex]
[tex]=\frac{720\times 10^{-9}\times 0.85}{0.22\times 10^{-3}}[/tex]
[tex]=2.78\times 10^{-3} \ m[/tex]
or,
[tex]=2.78 \ mm[/tex]
30.
the horizontal. The force needed to push the body up the plane is
A body of mass 20kg is pushed up a smooth plane inclined at an angle of 30° to
b. 200N c. 100N
d. 20N
a. ION
Answer:
b. 200N c. 100N
Explanation:
30.
the horizontal. The force needed to push the body up the plane is
If we convert a circuit into a current source with parallel load it is called?
Answer:
If we convert a circuit into a current source with parallel load it is called source transformation
Hey guys....
What is the advantage of a capacitor as it stores charge?
comparison between copper properties and aluminium properties
A 1000 kg weather rocket is launched straight up. The rocket motor provides a constant acceleration for 16 s, then the motor stops. The rocket altitude 20 s after launch is 6600 m. You can ignore any effects of air resistance.
Required:
a. What was the rocket's acceleration during the first 16s?
b. What is the rocket's speed as it passes through acloud 5100 m above the ground?
Answer:
a) a = 34.375 m / s², b) v_f = 550 m / s
Explanation:
This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.
a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed
v_f = [tex]\frac{x-x_1}{t}[/tex]
we substitute the values
v_f = [tex]\frac{ 6600 -x_1}{4}[/tex]
The initial part of the movement is carried out with acceleration
v_f = v₀ + a t
x₁ = x₀ + v₀ t + ½ a t²
the rocket starts from rest v₀ = 0 with an initial height x₀ = 0
x₁ = ½ a t²
v_f = a t
we substitute the values
x₁ = 1/2 a 16²
x₁ = 128 a
v_f = 16 a
let's write our system of equations
v_f = [tex]\frac{6600 - x_1}{4}[/tex]
x₁ = 128 a
v_f = 16 a
we substitute in the first equation
16 a = [tex]\frac{6600 -128 a}{4}[/tex]
16 4 a = 6600 - 128 a
a (64 + 128) = 6600
a = 6600/192
a = 34.375 m / s²
b) let's find the time to reach this height
x = ½ to t²
t² = 2y / a
t² = 2 5100 / 34.375
t² = 296.72
t = 17.2 s
We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part
v_f = 16 a
v_f = 16 34.375
v_f = 550 m / s
g as measured from the earth, a spacecraft is moving at speed .80c toward a second spacecraft moving at speed .60c back toward the first spacecraft. What is the speed of the first spacecraft as viewed from the second spacecraft
Answer:
the speed of the first spacecraft as viewed from the second spacecraft is 0.95c
Explanation:
Given that;
speed of the first spacecraft from earth v[tex]_a[/tex] = 0.80c
speed of the second spacecraft from earth v[tex]_b[/tex] = -0.60 c
Using the formula for relative motion in relativistic mechanics
u' = ( v[tex]_a[/tex] - v[tex]_b[/tex] ) / ( 1 - (v[tex]_b[/tex]v[tex]_a[/tex] / c²) )
we substitute
u' = ( 0.80c - ( -0.60c) ) / ( 1 - ( ( 0.80c × -0.60c) / c² ) )
u' = ( 0.80c + 0.60c ) / ( 1 - ( -0.48c² / c² ) )
u' = 1.4c / ( 1 - ( -0.48 ) )
u' = 1.4c / ( 1 + 0.48 )
u' = 1.4c / 1.48
u' = 0.9459c ≈ 0.95c { two decimal places }
Therefore, the speed of the first spacecraft as viewed from the second spacecraft is 0.95c
Forces applied in the opposite direction are
Added
Subtracted
Multiplied
Divided
Answer:
its number 2 one but i am not sure hope its right
Please helppppppp!!!!!!!!!!!!!!
Answer:
circuit breaker
Explanation:
A circuit breaker is a device used for electrical safety. It consists of a switch designed to protect an electrical circuit from damage that may result from heating due to overload in the circuit.
Its basic function is to interrupt current flow through its switch that consists of metal stripe which bends when it gets hot.
Fuse has similar action with circuit breaker, the only difference is that fuse can only be used once because it melts when it gets hot.
Therefore, the correct answer is "circuit breaker"
while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity of 8,32 meters per second take the speed of sound as 340 meters per second calculate frequency
Complete question:
while hunting in a cave a bat emits sounds wave of frequency 39 kilo hartz were moving towards a wall with a constant velocity of 8.32 meters per second take the speed of sound as 340 meters per second. calculate the frequency reflected off the wall to the bat?
Answer:
The frequency reflected by the stationary wall to the bat is 41 kHz
Explanation:
Given;
frequency emitted by the bat, = 39 kHz
velocity of the bat, [tex]v_b[/tex] = 8.32 m/s
speed of sound in air, v = 340 m/s
The apparent frequency of sound striking the wall is calculated as;
[tex]f' = f(\frac{v}{v- v_b} )\\\\f' = 39,000(\frac{340}{340 -8.32} )\\\\f' = 39978.29 \ Hz[/tex]
The frequency reflected by the stationary wall to the bat is calculated as;
[tex]f_s = f'(\frac{v + v_b}{v} )\\\\f_s = 39978.29(\frac{340 + 8.32}{340} )\\\\f_s = 40,956.56 \ Hz[/tex]
[tex]f_s\approx 41 \ kHz[/tex]
an object moves clockwise around a circle centered at the origin with radius m beginning at the point (0,). a. find a position function r that describes the motion of the object moves with a constant speed, completing 1 lap every s. b. find a position function r that describes the motion if it occurs with speed .
Answer:
Answer to An object moves clockwise around a circle centered at the origin with radius 6 m beginning at ... 6 M Beginning At The Point (0,6) B. Find A Position Function R That Describes The Motion If It Occurs With Speed E T A. R(t)= S The Motion Of The Object Moves With A Constant Speed, Completing 1 Lap Every 12 S.
Explanation:
A 0.50-m long solenoid consists of 500 turns of copper wire wound with a 4.0 cm radius. When the current in the solenoid is 22 A, the magnetic field at a point 1.0 cm from the central axis of the solenoid is
Answer: The magnetic field at a point 1.0 cm from the central axis of the solenoid is 0.0276 T.
Explanation:
Given: Length = 0.50 m
No. of turns = 500
Current = 22 A
Formula used to calculate magnetic field is as follows.
[tex]B = \mu_{o}(\frac{N}{L})I[/tex]
where,
B = magnetic field
[tex]\mu_{o}[/tex] = permeability constant = [tex]4\pi \times 10^{-7} Tm/A[/tex]
N = no. of turns
L = length
I = current
Substitute the values into above formula as follows.
[tex]B = \mu_{o}(\frac{N}{L})I\\= 4 \pi \times 10^{-7} Tm/A \times (\frac{500}{0.5 m}) \times 22\\= 0.0276 T[/tex]
Thus, we can conclude that magnetic field at a point 1.0 cm from the central axis of the solenoid is 0.0276 T.
16. Olympic ice skaters are able to spin at about 5 rev/s.
(a) What is their angular velocity in radians per second?
(b) What is the centripetal acceleration of the skater's nose it
it is 0.120 m from the axis of rotation?
Answer:
a) w = 31.4 rad / s, b) a = 118.4 m / s²
Explanation:
a) let's reduce to the SI system
w = 5 rev / s (2pi rad / 1 rev)
w = 31.4 rad / s
b) the expression for the centripetal acceleration is
a = v² / r
linear and angular variables are related
v = w r
we substitute
a = w² r
a = 31.4² 0.120
a = 118.4 m / s²
A long, straight wire lies along the zz-axis and carries a 3.90-AA current in the z z-direction. Find the magnetic field (magnitude and direction) produced at the following points by a 0.600 mmmm segment of the wire centered at the origin.
The question is incomplete. The complete question is :
A long, straight wire lies along the z-axis and carries a 3.90-A current in the + z-direction. Find the magnetic field (magnitude and direction) produced at the following points by a 0.600 mm segment of the wire centered at the origin.
A) x=2.00m,y=0, z=0
Bx,By,Bz = ? T
Enter your answers numerically separated by commas.
B) x=0, y=2.00m, z=0
C) x=2.00m, y=2.00m, z=0
D) x=0, y=0, z=2.00m
Solution :
The expression of the magnetic field using the Biot Savart's law is given by :
[tex]$d \vec B=\frac{\mu_0 I\vec{dl} \times \vec r}{4 \pi r^3}$[/tex]
a). The position vector is on the positive x direction.
[tex]$\vec r = (2 \ m) \ \hat i$[/tex]
[tex]$|r| = 2 \ m$[/tex]
The magnetic field is
[tex]$d \vec B=\frac{\mu_0 I\vec{dl} \times \vec r}{4 \pi r^3}$[/tex]
[tex]$d \vec B=\frac{4 \pi \times 10^{-7} \times 3.9 \times 0.6 \times 10^{-3} \times\hat k \times (2 ) \hat i }{4 \pi \times (2)^3}$[/tex]
[tex]$d \vec B=(5.85 \times 10^{-11} \ T)\hat j$[/tex]
The magnetic field is [tex]$(0, \ 5.85 \times 10^{-11} \ T, \ 0).$[/tex]
b). The position vector is in the positive y-direction.
[tex]$\vec r = (2 \ m) \ \hat j$[/tex]
[tex]$|r| = 2 \ m$[/tex]
The magnetic field is
[tex]$d \vec B=\frac{\mu_0 I\vec{dl} \times \vec r}{4 \pi r^3}$[/tex]
[tex]$d \vec B=\frac{4 \pi \times 10^{-7} \times 3.9 \times 0.6 \times 10^{-3} \times\hat k \times (2 ) \hat j }{4 \pi \times (2)^3}$[/tex]
[tex]$d \vec B=(5.85 \times 10^{-11} \ T)(-\hat{i})$[/tex]
The magnetic field is [tex]$(- 5.85 \times 10^{-11} \ T, \ 0, \ 0).$[/tex]
c). The position vector is :
[tex]$\vec r = (2)\hat i + (2)\hat j$[/tex]
[tex]$|\vec r| = \sqrt{(2)^2+(2)^2}$[/tex]
[tex]$=2.828 \ m$[/tex]
The magnetic field is
[tex]$d \vec B=\frac{\mu_0 I\vec{dl} \times \vec r}{4 \pi r^3}$[/tex]
[tex]$d \vec B=\frac{4 \pi \times 10^{-7} \times 3.9 \times 0.6 \times 10^{-3} \times\hat k \times ((2)\hat i + (2) \hat j) }{4 \pi \times (2.828)^3}$[/tex]
[tex]$=(4.13\times 10^{-11})\hat j+(4.13\times 10^{-11})(-\hat i)$[/tex]
The magnitude of the magnetic field is :
[tex]$|d\vec B|=\sqrt{(4.13\times 10^{-11})^2+(4.13\times 10^{-11})^2}$[/tex]
[tex]$=5.84 \times 10^{-11} \ T$[/tex]
Therefore, the magnetic field is [tex]$(-4.13 \times 10^{-11}\ T, \ 4.13 \times 10^{-11}\ T, \ 0 )$[/tex]
d). The position vector is in the positive y-direction.
[tex]$\vec r = (2 \ m) \ \hat k$[/tex]
[tex]$|r| = 2 \ m$[/tex]
The magnetic field is
[tex]$d \vec B=\frac{\mu_0 I\vec{dl} \times \vec r}{4 \pi r^3}$[/tex]
[tex]$d \vec B=\frac{4 \pi \times 10^{-7} \times 3.9 \times 0.6 \times 10^{-3} \times\hat k \times (2 ) \hat k }{4 \pi \times (2)^3}$[/tex]
= 0 T
The magnetic field is (0, 0, 0)
You want to produce a magnetic field of magnitude 5.50 x 10¹ T at a distance of 0.0 6 m from a long, straight wire's center. (a) What current is required to produce this field? (b) With the current found in part (a), how strong is the magnetic field 8.00 cm from the wire's center?
Answer:
(a) I = 1650000 A
(b) 4.125 T
Explanation:
Magnetic field, B = 5.5 T
distance, r = 0.06 m
(a) Let the current is I.
The magnetic field due to a long wire is given by
[tex]B =\frac{\mu o}{4\pi }\frac{2 I}{r}\\5.5= 10^{-7}\times \frac{2\times I}{0.06}\\I =1650000 A[/tex]
(b) Let the magnetic field is B' at distance r = 0.08 m.
[tex]B =\frac{\mu o}{4\pi }\frac{2 I}{r}\\B = 10^{-7}\times \frac{2\times 1650000}{0.08}\\B'= 4.125 T[/tex]
One product of radioactive decay is Alpha Radiation, which consists of Hydrogen nuclei composed of one proton and no neutrons.
a. True
b. False
Answer:
False
Explanation:
The alpha decay or alpha radiation is one type of radioactive decay. What is emitted is an alpha particle which is helium nucleus and not the hydrogen nucleus. The alpha particle is made up of two protons as well as two neutrons. This is the helium nucleus.
Therefore the right answer to this question is false.
Two long, straight wires are fixed parallel to one another a distance do apart. The wires carry equal constant currents 1, in the same direction. The attractive magnetic force per unit length between them if f = F/L. What is the force per unit length between the wires if their separation is 2d, and each carries current 2I0?
A. f/4
B. f/2
C. 3f/2
D.) 2f
Answer:
Option D
Explanation:
From the question we are told that:
The attractive magnetic force per unit length as
[tex]f = F/L[/tex]
Separation Distance [tex]x=2d[/tex]
Generally the equation for Magnetic force between two current carrying wire is mathematically given by
[tex]\frac{F}{\triangle l}=\frac{\mu_0I_1I_2}{\mu \pi x}[/tex]
[tex]\frac{F}{\triangle l }=\frac{I_1I_2}{ x}[/tex]
Where
[tex]x=2r[/tex]
And
[tex]I_1=I_2=>2I[/tex]
Then
[tex]\frac{F}{\triangle l}=>\frac{2*2}{2}*f[/tex]
[tex]\frac{F}{\triangle l}=>2f[/tex]
Therefore s the force per unit length between the wires if their separation is 2d
[tex]\frac{F}{\triangle l}=>2f[/tex]
Option D
How much energy must be added to a 1-kg piece of granite with a specific
heat of 600 J/(kg°C) to increase its temperature from 20° C to 100° C?
A. 48,000 J
B. 4,800 J
C. 1,200,000 J
D. 60,000 J
Answer: 48,000 J
Explanation: i just did it
An electron travels 1.49 m in 7.4 µs (microsecWhat is its speed if 1 inch = 0.0254 m? Answer in units of in/min.
Explanation:
Write what you know
Speed = Distance / Time
micro- = 10^-6
write your conversions as fractions
1 in / 0.0254 m
1 min / 60 sec
First convert time to regular seconds
7.4 x 10^-6 seconds
Use Velocity
1.49m / (7.4 x 10^-6) s
We've written our conversions in fractions because units cancel out just like numbers
[tex] \frac{1.49m}{7.4 \times {10}^{ - 6} } \times \frac{1in}{0.0254m} \times \frac{60sec}{1min} [/tex]
Multiply all the fractions accross and youll have your answer
PLEASE HELPPP MEEE :((
Physics part 2
These the other questions 14 - 17
Answer:
bvihobonlnohovicjfufufufucvkvkvvjcufufydyfuvi
A person carries a plank of wood 1.6 m long with one hand pushing down on it at one end with a force F1 and the other hand holding it up at 43 cm from the end of the plank with force F2. If the plank has a mass of 13.7 kg and its center of gravity is at the middle of the plank, what is the force F1
Answer: [tex]115.52\ N[/tex]
Explanation:
Given
Length of plank is 1.6 m
Force [tex]F_1[/tex] is applied on the left side of plank
Force [tex]F_2[/tex] is applied 43 cm from the left end O.
Mass of the plank is [tex]m=13.7\ kg[/tex]
for equilibrium
Net torque must be zero. Taking torque about left side of the plank
[tex]\Rightarrow mg\times 0.8-F_2\times 0.43=0\\\\\Rightarrow F_2=\dfrac{13.7\times 9.8\times 0.8}{0.43}\\\\\Rightarrow F_2=249.78\ N[/tex]
Net vertical force must be zero on the plank
[tex]\Rightarrow F_1+W-F_2=0\\\Rightarrow F_1=F_2-W\\\Rightarrow F_1=249.78-13.7\times 9.8\\\Rightarrow F_1=115.52\ N[/tex]
Your friend has been given a laser for her birthday. Unfortunately, she did not receive a manual with it and so she doesn't know the wavelength that it emits. You help her by performing a double-slit experiment, with slits separated by 0.36 mm. You find that the two m n = 2 bright fringes are 5.5 mm apart on a screen 1.6 m from the slits.
a. What is the wavelength the light emits?
b. What is the distance between the two n = 1 dark fringes?
Answer:
a) the wavelength that the light emits is 6.1875 × 10⁻⁷ m
b) the distance between the two n = 1 dark fringes is 5.5 × 10⁻³ m
Explanation:
Given the data in the question;
separation between two slits d = 0.36 mm = 0.00036 m
Separation between two adjacent fringes β = 5.5 mm = 0.0055 m
Distance of screen from slits D = 1.6 m
n = 2
a) the wavelength the light emits;
Using the formula;
β = (nD/d)λ
To find wavelength, we make λ the subject of formula;
βd = nDλ
λ = βd / nD
so we substitute
λ = ( 0.0055 m × 0.00036 m ) / ( 2 × 1.6 m )
λ = 0.00000198 / 3.2
λ = 6.1875 × 10⁻⁷ m
Therefore, the wavelength that the light emits is 6.1875 × 10⁻⁷ m
b) the distance between the two n = 1 dark fringes;
To find the distance between the two n = 1 dark fringes, we use the following formula;
y[tex]_m[/tex] = 2nλD / d
given that n = 1, we substitute
y[tex]_m[/tex] = ( 2 × 1 × ( 6.1875 × 10⁻⁷ m ) × 1.6 m ) / 0.00036 m
y[tex]_m[/tex] = 0.00000198 / 0.00036
y[tex]_m[/tex] = 0.0055 m
y[tex]_m[/tex] = 5.5 × 10⁻³ m
Therefore, the distance between the two n = 1 dark fringes is 5.5 × 10⁻³ m
A proton is accelerated from rest through a potential difference V0 and gains a speed v0. If it were accelerated instead through a potential difference of 3V0, what speed would it gain? Group of answer choices
Answer:
[tex]v_{0,new} = v0\sqrt{}2[/tex]
Explanation:
Initial work done on the proton is given by, [tex]\DeltaW0 = q V_o[/tex]
we know that, [tex]\DeltaW = \DeltaK.E[/tex]
[tex]qV0 = (1/2) m v_0^2[/tex]
[tex]v_0 = \sqrt{}2 q V_0 / m[/tex] { eq.1 }
If it were accelerated instead through a potential difference of 2V0, then it would gain a speed will be given as :
using the above formula, we have
[tex]v_{0,new} = \sqrt{}2 q (2V0) / m[/tex]
[tex]v_{0,new} = \sqrt{}4 q V0 / m[/tex]
[tex]v_{0,new} = v0\sqrt{}2[/tex]
Suppose that the position of a particle is given by s=f(t)=5t3+6t+9. (a) Find the velocity at time t.
This question is incomplete, the complete question is;
Suppose that the position of a particle is given by s=f(t)=5t³ + 6 t+ 9.
(a) Find the velocity at time t.
(b) Find the velocity at time t=3 seconds
Answer:
a) the velocity at time t is ( 15t² + 6 ) m/s
b) Velocity at time t=3 seconds is 141 m/s
Explanation:
Give the data in the question;
position of a particle is given by;
s = f(t) = 5t³ + 6t + 9
Velocity at t;
we differentiate with respect to t
so
V(t) = f'(t) = d/dt ( 5t³ + 6t + 9 )
V(t) = f(t) = 5(3t²) + 6(1) + 0 )
V(t) = f(t) = ( 15t²+6 ) m/s
Therefore, the velocity at time t is 15t²+6 m/s
b) Velocity at t = 3 seconds
V(t) = f(t) = ( 15t²+6 ) m/s
we substitute
V(3) = ( 15(3)² + 6 ) m/s
V(3) = ( (15 × 9) + 6 ) m/s
V(3) = ( 135 + 6 ) m/s
V(3) = 141 m/s
Therefore, Velocity at time t=3 is 141 m/s
a body thrown vertically upwards from grounf with inital vel 40m/s then time taken by it to reach max hieght is?
Answer:
t = 4.08 s
Explanation:
if the body is thrown upward, it has negative gravity. Knowing through the International System that the earth's gravity is 9.8 m/s²
Data:
Vo = 40 m/sg = -9.8 m/s²t = ?Use formula:
[tex]\boxed{\bold{t=\frac{-(V_{0})}{g}}}[/tex]Replace and solve:
[tex]\boxed{\bold{t=\frac{-(40\frac{m}{s})}{-9.8\frac{m}{s^{2}}}}}[/tex][tex]\boxed{\boxed{\bold{t=4.08\ s}}}[/tex]Time taken by it to reach max height is 4.08 seconds.
Greetings.
What is measured by the change in velocity of a moving object?
Answer:
acceleration is measured
as a mercury atom absorbs a photon of energy as electron in the atom changes from energy level B to energy level E. calculate the frequency of the absorb photon.
Answer:
2.00x 10 14th Hz
Explanation:
Answer:
2.99 x 10^14 Hz
Explanation:
E photon= hf (you have to solve for f)
f= E photon/h
f= 1.98 x 10^-19 J / 6.63 x 10^-34 J x s
f=2.99 x 10^14 Hz
A vacuum gauge connected to a tank reads 30.0 kPa. If the local atmospheric pressure is 13.5 psi, what is the absolute pressure in units of psi, with 3 sig figs
Answer:
[tex]P_a=17.85psi[/tex]
Explanation:
From the question we are told that:
Tank Pressure [tex]P_t=30.0kpa[/tex]
Atmospheric Pressure [tex]P_a=13.5 psi[/tex]
Where
[tex]1kpa=0.148psi[/tex]
Therefore
[tex]30kpa=4.35psi[/tex]
Generally the equation for Absolute pressure [tex]P_a[/tex] is mathematically given by
[tex]P_a=13.5+4.35[/tex]
[tex]P_a=17.85psi[/tex]
Question 11 of 22
A horse of mass 180 kg gallops at a speed of 8 m/s. What is the momentum
of the horse?
Answers
1440
22.5
845
1955
Momentum = (mass) x (speed)
If you work the problem in the same units as the given data, then you get the momentum in units of kilogram-meters per second, and your horse has 1,440 of them.
Answer:
A
Explanation:
1440 kg*m/s