Answers
Answer:
The value of [tex]\theta=63.43^0[/tex]
Explanation:
Range of projectile [tex]\mathrm{R}=\frac{\mathrm{u}^{2} \cdot \sin 2 \theta}{\mathrm{g}}[/tex]
Height of projectile [tex]\mathrm{h}=\frac{\mathrm{u}^{2} \cdot \sin ^{2} \theta}{2 \cdot \mathrm{g}}[/tex]
Here we have
[tex]R=2h\\\\\frac{\mathrm{u}^{2} \cdot \sin 2 \theta}{\mathrm{g}}=2\frac{\mathrm{u}^{2} \cdot \sin ^{2} \theta}{2 \cdot \mathrm{g}}\\\\\sin 2 \theta=\sin ^{2} \theta\\\\2\sin \theta\cos \theta=\sin ^{2} \theta\\\\\tan \theta=2\\\\\theta=63.43^0[/tex]
Range be R and height be h
[tex]\boxed{\sf R=\dfrac{u^2sin2\theta}{g}}[/tex]
[tex]\boxed{\sf h=\dfrac{u^2sin^2\theta}{2g}}[/tex]
u=initial velocitytheta is angle of projection.g=acceleration due to gravityATQ
[tex]\\ \sf\longmapsto R=2h[/tex]
[tex]\\ \sf\longmapsto \dfrac{u^2sin2\theta}{g}=\dfrac{2u^2sin^2\theta}{2g}[/tex]
Cancelling required ones[tex]\\ \sf\longmapsto sin^2\theta=sin2\theta[/tex]
sin2O=2sinOcosO[tex]\\ \sf\longmapsto sin^2\theta=2sin\theta cos\theta [/tex]
[tex]\\ \sf\longmapsto \dfrac{sin^2\theta}{sin\theta cos\theta=2[/tex]
[tex]\\ \sf\longmapsto \dfrac{sin\theta}{cos\theta}=2[/tex]
[tex]\\ \sf\longmapsto tan\theta=2[/tex]
[tex]\\ \sf\longmapsto \theta=tan^{-1}(2)[/tex]
[tex]\\ \sf\longmapsto \theta=63.4°[/tex]
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6. A 3.4-kg bucket of water is attached to a 1.0-m rope. The bucket is swung in a circle at a speed
of 10.0-m/s.
a.
If the rope can only tolerate 400-N of force, what is the maximum speed the bucket can
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Answers
Answer:
Explanation:
hope this helps
The maximum speed of the bucket can experience before the rope snaps if A 3.4-kg bucket of water is attached to a 1.0-m rope. The bucket is swung in a circle at a speed of 10.0-m/s is 10.85 m / s.
What is force?Force is the influence of either pull or pushes in the body. Basically, gravitation forces, nuclear forces, and friction forces are the types of forces. For e.g. when the wall is hit by a hand then a force is exerted by the hand on the wall as well as the wall also exerts a force on the hand. There are different laws given to Newton to understand force.
Newton is a unit of force used by physicists that is part of the International System (SI). The force required to move a body weighing one kilogram one meter per second is known as a newton.
Given:
The mass of the bucket, m = 3.4 kg,
The length of the rope, r = 1 m,
The speed of the bucket, v = 10 m / s,
The force of the rope, F = 400 N,
Calculate the maximum speed by the formula given below,
[tex]F_c = mv^2 / r[/tex]
400 = 3.4 v² / 1
v² = 400 / 3.4
v² = 117.6
v = 10.85 m / s
Therefore, the maximum speed of the bucket can experience before the rope snaps if A 3.4-kg bucket of water is attached to a 1.0-m rope. The bucket is swung in a circle at a speed of 10.0-m/s is 10.85 m / s.
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Resolve the weight of the box to find the component of the weight acting parallel to the slope.
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30
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Answer:
here's your answer below
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sorry something went wrong
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Which design of the walls would be most desirable help avoid echoes?
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Answers
When the mass of the cylinder increases from 3. 0 kg to 6. 0 kg, the amount of heat generated in the system increases by a factor of 2.
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Answer:
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Answer:
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Hi there!
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Considering the ideal gas law, 116.06 moles of are needed to fill a 2000 L weather balloon at 210 K and 1 atm pressure.
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Applying Charles law
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Answers
Answer:
they are machined with shape characteristics which maximize the fatigue life of a metal.
Explanation:
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Answers
Time taken by the skydiver to land on the ground is 4.57 seconds
The applied force, F = 700 N
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To calculate the time taken by the skydiver to fall to the ground will be calculated using the formula
[tex]F=\frac{mv}{t}[/tex]
Substitute F = 700, m = 80, and v = 40 to solve for t
[tex]700=\frac{80(40)}{t}\\\\700t=3200\\\\t=\frac{3200}{700}\\\\t=4.57 s[/tex]
Time taken by the skydiver to land on the ground is 4.57 seconds
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Answers
Answer:
Flow is proportional to change in pressure and inversely proportional to resistance.