Answer:
Because shoes protect our feet from some of the most harmful platforms
Gravel has some small pebbles on it sometimes (or other sharp objects)
Gravel is pretty hard.
How do unbalanced forces acting on an object affect its motion when the object is at rest? What if it is moving?
Answer:
It pushes it because an unbalanced force is pushing more newtons than something that isn't even moving. Even if it is moving, it depends which side is pushing/pulling the most force.
Explanation:
Answer:
It pushes it because an unbalanced force is pushing more newtons than something that isn't even moving. Even if it is moving, it depends which side is pushing/pulling the most force.
This is a short question can anyone help me please
Thank you
Picture Above
Answer:
I thinks it's
deficit spending
Explanation:
cause When a government spends more than it collects in taxes, it is said to have a budget deficit.
Elevator is accelerating upward 3.5 M/S2 and has a mass of 300 KG. The force of gravity is 2940 N. What is the tension force pulling elevator up?
Answer:
T = 3990 N
Explanation:
The free body diagram for the elevator consists of a tension force pointing up, and its weight pointing down. So the elevator's net force is:
F = T - 2940N
ad at the same time, using Newton's second law, we have that this net force should equal the elevator's mass (300 kg) times its acceleration (a):
T - 2940N = 300kg (3.5m/s^2)
then
T = 2940 N + 1050 N
T = 3990 N
Which statement best describes a characteristic of gases?
Gases can be compressed, or squeezed together.
The particles of gases are packed close together.
The particles of gases spread our vertically instead of horizontally.
Gases have a definite shape and volume.
Answer:
A
Explanation:
I'm pretty sure it's A. That's the one that makes the most sense and checks out
Answer:
Just here to confirm that it is A
Explanation:
Mr. Jones starts from rest and begins to accelerate straight to the bathroom at a rate of 0.5 m/s for 10 seconds. What kind of motion is this
A. linear
b centripetal
c.free fall
d projectile
Answer:
A. linear
Explanation:
because they are going in a straight line
Approximating Venus's atmosphere as a layer of gas 50 km thick, with uniform density 21 kg/m3, calculate the total mass of the atmosphere.Express your answer using two significant figures.m venus atmosphere = ____ kg
Answer:
m = 4.9 10⁸ kg
Explanation:
The expression for the density is
ρ = m / V
m = ρ V
the volume of the atmosphere is the volume of the sphere of the outer layer of the atmosphere minus the volume of the plant
V = V_atmosphere - V_planet
V = 4/3 π R_atmosphere³ - 4/3 π R_venus³
V = 4/3 π (R_atmosphere³ - R_venus³
)
the radius of the planet is R_venus = 6.06 10⁶ m.
The radius of the outermost layer of the atmosphere
R_atmosphere = 50 10³ + R_ venus = 50 10³ + 6.06 10⁶
R_atmosphere = 6.11 10⁶ m
let's find the volume
V = 4/3 pi [(6,11 10⁶)³ - (6,06 10⁶)³]
V = 23,265 10⁶ m³
let's calculate the mass
m = 21 23,265 10⁶
m = 4.89 10⁸ kg
with two significant figurars is
m = 4.9 10⁸ kg
F = 5 Newtons
W = 75 Joules
d = ?
ANSWER
What physical property does the symbol I_enclosed in problem 5 represent? a. The current along the path in the same direction as the magnetic field b. The current in the path in the opposite direction from the magnetic field c. The total current passing through the loop in either direction d. The net current through the loop
Answer:
C
Explanation:
Current passing through the loop in either direction
A uniform electric field has a magnitude of 10 N/C and is directed upward. A charge brought into the field experiences a force of 50 N downward. The charge must be:_______.
Answer:
q = 5 C
Explanation:
The electric field is defined as the force experienced by a unit charge when it is brought into the field. Hence, the formula used to find the electrical field is given as follows:
E = F/q
where,
E = Electric Field Magnitude = 10 N/C
F = Force Experienced by the test charge = 50 N
q = Magnitude of the Charge = ?
Therefore,
10 N/C = 50 N/q
q = 50 N/(10 N/C)
Therefore,
q = 5 C
Net force causes motion
Answer:
yes
Explanation:
An electron is accelerated from rest through a potential difference of 300V. it then passes through a uniform 0.001-T magnetic field, oriented perpendicular to the electrons velocity. what is the magnitude of the magnetic force on the electron?
Given :
Potential difference, V = 300 V.
Magnetic Field, B = 0.001 T.
To Find :
The magnitude of the magnetic force on the electron.
Solution :
We know, for perpendicular orientation, force is given by :
[tex]F = qVB\\\\F = 1.6 \times 10^{-19} \times 300\times 0.001\ N\\\\F = 4.8\times 10^{-20}\ N[/tex]
Hence, this is the required solution.
A parallel-plate capacitor with circular plates of radius 40 mm is being discharged by a current of 6.0 A. At what radius (a) inside and (b) outside the capacitor gap is the magnitude of the induced magnetic field equal to 75% of its maximum value?(c) What is that maximum value?
Answer:
A) r = 0.03 m
B) r = 0.0533 m
C) B_max = 0.00003 T
Explanation:
Formula for magnetic field inside the capacitor when it is parallel to the length element is;
B_in = (μ_o•I•r/(2πR²)
Formula for maximum magnetic field is;
B_max = (μ_o•I/(2πR)
Formula for magnetic field outside the capacitor is; B_out = (μ_o•I/(2πr)
A) Magnetic field inside the capacitor is gotten from our first equation above;
B_in = (μ_o•I•r/R²)
Since we want to find the radius at which the magnitude of the induced magnetic field equal to 75% or 0.75 of its maximum value.
Thus;
B_in = 0.75B_max
(μ_o•I•r/(2πR²) = 0.75((μ_o•I/(2πR))
μ_o•I and 2πR will cancel out to give;
r/R = 0.75
r = 0.75R
We are given R = 40 mm = 0.04 m
r = 0.75 × 0.04
r = 0.03 m
B) magnetic field outside the capacitor is; B_out = (μ_o•I/(2πr)
Thus for the magnitude of the induced magnetic field equal to 75% or 0.75 of its maximum value:
B_out = 0.75B_max
(μ_o•I/(2πr) = 0.75((μ_o•I/(2πR))
μ_o•I and 2π will cancel out to give;
1/r = 0.75/R
r = R/0.75
r = 0.04/0.75
r = 0.0533 m
C) B_max = μ_o•I/(2πR)
μ_o is a constant known as vacuum of permeability with a value of 4π × 10^(-7) T.m/A
Thus;
B_max = (4π × 10^(-7) × 6)/(2π × 0.04)
B_max = 0.00003 T
When a mass of a cart is 10 kg, and an applied force is 5 N, The acceleration of the cart is
5.0 m/s2
2.0 m/s2
0.5 m/s2
0.2 m/s2
Answer:
0.5 m/s2
Explanation:
F = ma
5 = 10a
a = 5/10
a =0.5
A 8.45μC particle with a mass of 6.15 x 10^-5 kg moves perpendicular to a 0.493-T magnetic field in a circular path of radius 34.1 m. How much time will it take for the particle to complete one orbit?
a. 92.7 s
b. 0.0927 s
c. 9.27 s
d. 927 s
This question is incomplete, the complete question is;
A 8.45μC particle with a mass of 6.15 x 10⁻⁵ kg moves perpendicular to a 0.493-T magnetic field in a circular path of radius 34.1 m.
How much time will it take for the particle to complete one orbit?
a. 92.7 s
b. 0.0927 s
c. 9.27 s
d. 927 s
Answer:
it will take 92.7 seconds for the particle to complete one orbit.
Option a) 92.7 s is the correct option
Explanation:
Given that;
mass m = 6.15 x 10⁻⁵ kg
q = 8.45μC = 8.45 × 10⁻⁶ C
B = 0.493
we know that
Time period T = 2πr / V
where r = mv/qB
so T = 2πm/qB
we substitute
T = (2 × 3.14 × 6.15 x 10⁻⁵) / ( 8.45 × 10⁻⁶ × 0.493)
T = 0.0003862 / 0.000004165
T = 92.7 sec
Therefore it will take 92.7 seconds for the particle to complete one orbit.
Option a) 92.7 s is the correct option
What is the volume of a box if he has Length=7 cm Width=5cm , Height=10cm ?
Answer:
Volume of Cuboid = Height*Width*Length
Explanation:
Volume of Cuboid = 10*5*7
= 350 cu² cm
Answer:
Diagram:-[tex]\setlength{\unitlength}{0.74 cm}\begin{picture}\thicklines\put(5.6,5.4){\bf A}\put(11.1,5.4){\bf B}\put(11.2,9){\bf C}\put(5.3,8.6){\bf D}\put(3.3,10.2){\bf E}\put(3.3,7){\bf F}\put(9.25,10.35){\bf H}\put(9.35,7.35){\bf G}\put(3.5,6.1){\sf 5cm}\put(7.7,6.3){\sf 7cm}\put(11.3,7.45){\sf 10cm}\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(4,7.3){\line(1,0){5}}\put(4,10.3){\line(1,0){5}}\put(9,10.3){\line(0,-1){3}}\put(4,7.3){\line(0,1){3}}\put(6,6){\line(-3,2){2}}\put(6,9){\line(-3,2){2}}\put(11,9){\line(-3,2){2}}\put(11,6){\line(-3,2){2}}\end{picture}[/tex]
Required Answer:-It is a cuboid
where
length =l=7cmwidth=b=5cmheight =h=10cmAs we know that in a cuboid
[tex]{\boxed{\sf Volume=lbh}}[/tex]
Substitute the values[tex]{:}\longrightarrow[/tex][tex]\sf Volume =7×5×10 [/tex]
[tex]{:}\longrightarrow[/tex][tex]\sf Volume=35×10 [/tex]
[tex]{:}\longrightarrow[/tex][tex]\sf Volume=350cm^3 [/tex]
A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force) acts on her. The student has a mass of 71.0 kg, and the height of the water slide is 12.7 m. If the kinetic frictional force does -5.10 103 J of work, how fast is the student going at the bottom of the slide
Answer:
[tex]10.27m/s[/tex]
Explanation:
Given data
work W= -5.10 10^3 J
mass m= 71kg
final height of slide h2= 12.7m
initial height of slide h1=0m
initial velocity v1= 0m/s
final velocity v2=?
Step two:
required
Final velocity
The work-energy theorem is expressed as'
[tex]W=1/mv_2^2 +mgh_2-(1/mv_1^2+mgh_1)[/tex]
make V2 subject of formula we have final speed
[tex]v_2=\sqrt{\frac{2W}{m}+v_1^2-2g(h_1-h_2) } \\\\[/tex]
substitute our given data we have
[tex]v_2=\sqrt{\frac{2*(-5.1*10^3)}{71}+0^2-2*9.81(12.7) } \\\\v_2=\sqrt{143.66-249.174 } \\\\v_2=\sqrt{105.514 } \\\\v_2=10.27m/s[/tex]
The student going at 10.27m/s
Any five physics problems
Explanation:
There are still some questions beyond the Standard Model of physics, such as the strong CP problem, neutrino mass, matter–antimatter asymmetry, and the nature of dark matter and dark energy.
A 500-N box is at rest on the floor. Dennis Elbo makes several
attempts to move the box, pushing against the box with varying
amounts of horizontal force. Yet the box never does move. In this
situation, the amount of static friction force experienced by the box
Select all that apply.
-
0 is 500 N
O is equal to the force with which Dennis exerts on the
box
has an upper limit and Dennis O has not yet exceeded the upper limit
Ois always the coefficient of friction multiplied by the normal force value
Answer:
Select the second and the third options you listed.
Explanation:
Select the answer options:
"is equal to the force with which Dennis exerts on the box."
and
"has an upper limit and Dennis has not yet exceeded the upper limit."
In fact, this upper limit of the static friction force is the product of the coefficient of static friction ([tex]\mu[/tex]) times the weight of the box.
What percentage of an iron anchor’s weight will be supported by buoyant force when submerged in salt water?
Answer:
0.87
Explanation:
To solve this, we use the principle of Archimedes. Archimedes Principal of flotation states that "the buoyant force of an object is equal to the total weight of the fluid it displaces."
In the attachment, I stated the mathemacal formula, of which
F(B) = The buoyant force
w(fl) = The weight of the salt water displaced
p(iron) = density of iron
p(salt) = density of the salt water = 1025 kg/m³
F' = weight of the iron in air
F = weight of the iron in salt water
p(man) = density of man = 7680 kg/m³
The rest are the easy calculations done by substituting the values
A crossbow is fired horizontally off a cliff with an initial velocity of 15 m/s. If the arrow takes 4s to hit the ground, what is the range of the projectile?
Answer:
The range of the projectile is 60 m
Explanation:
Horizontal Motion
When an object is thrown horizontally with a speed vo from a height h, it describes a curved path ruled exclusively by gravity until it eventually hits the ground.
The horizontal component of the velocity is always constant because no acceleration acts in that direction, thus:
[tex]v_x=v_o[/tex]
The vertical component of the velocity changes in time because gravity makes the object fall at increasing speed given by:
[tex]v_y=g.t[/tex]
The horizontal distance is calculated as a constant speed motion:
[tex]x = v_x.t[/tex]
Knowing the crossbow is fired horizontally at vo=vx=15 m/s and it takes t=4 s to hit the ground, thus the range of the projectile is:
x = 15*4 = 60
The range of the projectile is 60 m
when you stir a cup of tea the floating clip collect at the centre of the Cup rather than in the outer Rim why
Answer:
hsvshxansjusjsnwjwisks
Explanation:
When we stir a cup of tea we create a force in the center which pulls out all the particles towards it this is the basic reason for collection of tea leaves at the center of the cup rather than at the rim of the cup, it is similar to the the case of tornado where it takes all the particles present on it way to its ..
A bullet of mass 4.2 g strikes a ballistic pendulum of mass 1.0 kg. The center of mass of the pendulum rises a vertical distance of 18 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.
Answer:
449.020m/s
Explanation:
We have been provided with the following information:
Mass of bullet = Mb = 4.2g
Mass of pendulum = 1.0 = Md
Vertical distance = h = 18cm
Gravitational force = g = 9.8
We have kinetic energy converted to potential energy for the entire system
1/2(Mb+Md)V² = (Mb + Md)gh
We have V as the speed of system during collision
Mbv = (Mb+Md)V
We divide through by Mb
v = (Mb+Md/Mb)√2gh
4.2x10^-3 = 0.0042
18cm = 0.18m
(0.0042+1.0/0.0042)√2x9.8x0.18
= 1.0042/0.0042√3.528
= 239.095x1.878
= 449.020m/s
This is the answer to the question
Thank you!
A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns. They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) (4.00 m/s2 ) .At that instant and in unit-vector notation, what is the acceleration of the wallet
Complete Question:
A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns.
They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) i + (4.00 m/s2 ) j .At that instant and in unit-vector notation, what is the acceleration of the wallet
Answer:
aw = 3 i + 6 j m/s2
Explanation:
Since both objects travel in uniform circular motion, the only acceleration that they suffer is the centripetal one, that keeps them rotating.It can be showed that the centripetal acceleration is directly proportional to the square of the angular velocity, as follows:[tex]a_{c} = \omega^{2} * r (1)[/tex]
Since both objects are located on the same radial line, and they travel in uniform circular motion, by definition of angular velocity, both have the same angular velocity ω.∴ ωp = ωw (2)
⇒ [tex]a_{p} = \omega_{p} ^{2} * r_{p} (3)[/tex]
[tex]a_{w} = \omega_{w}^{2} * r_{w} (4)[/tex]
Dividing (4) by (3), from (2), we have:[tex]\frac{a_{w} }{a_{p}} = \frac{r_{w} }{r_{p}}[/tex]
Solving for aw, we get:[tex]a_{w} = a_{p} *\frac{r_{w} }{r_{p} } = (2.0 i + 4.0 j) m/s2 * 1.5 = 3 i +6j m/s2[/tex]
A kangaroo jumps straight up to a vertical height of 1.45 m. How long was it in the air before returning to Earth?
Answer:
1.08 s
Explanation:
From the question given above, the following data were obtained:
Height (h) reached = 1.45 m
Time of flight (T) =?
Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:
Height (h) = 1.45 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
1.45 = ½ × 9.8 × t²
1.45 = 4.9 × t²
Divide both side by 4.9
t² = 1.45/4.9
Take the square root of both side
t = √(1.45/4.9)
t = 0.54 s
Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).
Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:
Time (t) taken to reach the height = 0.54 s
Time of flight (T) =?
T = 2t
T = 2 × 0.54
T = 1.08 s
Therefore, it will take the kangaroo 1.08 s to return to the earth.
find the fundamental units involved in derived units
newton
watt
joule
pascal
cubic meter
9ycy8c8t 7f fixfuozofuxt8lsrupsurpaurae6pUeoUe6eoUeFipzuroz6d0, 7d0z6e0z7e0zurpz6e0z
Explanation:
force newton N - m·kg·s-2
pressure, stress pascal Pa N/m2 m-1·kg·s-2
energy, work, quantity of heat joule J N·m m2·kg·s-2
power, radiant flux watt W J/s m2·kg·s-3
volume cubic meter m3
The sound intensity at 4 m from a source is 100 W/me. What is the intensity of the sound at 12 m away from the source ?
Answer:
Intensity at 12 meters will be 11.11 W/m^2
Explanation:
Recall that the intensity of sound is inversely proportional to the square of the distance from the source. Therefore, if at 4 m the intensity is 100 W/m^2
we have: 100 W/m^2 = k/16 and therefore, k = 1600 W
Then the intensity (I) at 12 m will be:
I = k/12^2 = 1600/144 W/m^2 = 11.11 W/m^2
what is gathering and analyzing information about an object without physical contact with the object
Answer:
Remote Sensing
Explanation:
Which of the following is NOT a type of variable?
A. Burette
B. Ordered
C. Continuous
D. Categoric
Answer:
A
Explanation:
because its not trust me :)
Objects falling through air are slowed by the force of air resistance. Which objects were slowed the most by air resistance?
Answer:
This question is incomplete
Explanation:
This question is incomplete. However, it should be noted that when objects of different sizes fall in absence of air resistance, the objects will get to the ground at the same time. But with the presence of air resistance, the heaviest object gets to the ground first; meaning it has the least air resistance while the lightest object will arrive at the ground last because it has the greatest air resistance and is slowed down the most by the air resistance.
Thus, the lightest object in the completed question is the answer.
What is the average power consumption in watts of an appliance that uses 4.69 kW · h of energy per day?
Answer:
The average power is [tex]P = 195 .42 \ W[/tex]
Explanation:
From the question we are told that
The energy of the appliance is [tex]E = 4.69 \ kWh = 4.69 *10^{3} \ \ Wh[/tex]
The time considered is [tex]t = 1 \ day = 24 \ hours[/tex]
Generally the average power consumption is mathematically represented as
[tex]P = \frac{E}{t}[/tex]
=> [tex]P = \frac{4.69 *10^{3}}{24 }[/tex]
=> [tex]P = 195 .42 \ W[/tex]