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Elis [28]
3 years ago
12

A cheetah starts from rest and accelerates after a gazelle at a rate of 6.5 meters per second2for 3.0 seconds. Calculate the che

etah’s speed at the end of these 3.0 seconds.
Physics
1 answer:
Pepsi [2]3 years ago
6 0

Answer:

the speed of the cheetah at the end of the 3 seconds is: 19.5 m/s

Explanation:

Let's use the equation that relates speed with acceleration:

vf = vi + a * t

where vf stands for final velocity, vi stands for initial velocity, a for acceleration, and t for the time acceleration is applied. Then, in our case we have:

vf = 0 + 6.5 (3)

vf = 19.5 m/s

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(a) When a battery is connected to the plates of a 8.00-µF capacitor, it stores a charge of 48.0 µC. What is the voltage of the
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Answer:

a.6 V

b.24 V

Explanation:

We are given that

a.C=8\mu F=8\times 10^{-6} F

1\mu =10^{-6}

Q=48\mu C=48\times 10^{-6} C

We know that

V=\frac{Q}{C}

Using the formula

V=\frac{48\times 10^{-6}}{8\times 10^{-6}}=6 V

b.Q=192\mu C=192\times 10^{-6} C

V=\frac{192\times 10^{-6}}{8\times 10^{-6}}=24 V

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3 years ago
Describe what it means to view something from a frame of reference.Give an example to illustrate your explanation.
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2 years ago
a ballistic pendulum is used to measure the speed of high-speed projectiles. A 6 g bullet A is fired into a 1 kg wood block B su
Galina-37 [17]

Answer:

(a) v-bullet = 399.04 m/s

(b) I = 2.38 kg m/s

(c) T = 2.59 N

Explanation:

(a) To calculate the initial speed of the bullet, you first take into account that the kinetic energy of both wood block and bullet, just after the bullet impacts the block, is equal to the potential gravitational energy of block and bullet when the cord is at 60° respect to the vertical.

The potential energy is given by:

U=(M+m)gh       (1)

U: potential energy

M: mass of the wood block = 1 kg

m: mass of the bullet = 6g = 6.0*10^-3 kg

g: gravitational constant = 9.8m/s^2

h: distance to the ground

The distance to the ground is calculate d by using the information about the length of the cord and the degrees of the cord respect to the vertical:

h=l-lsin\theta\\\\h=2.2m-2,2m\ sin60\°=0.29m

The potential energy is:

U=(1kg+6*10^{-3}kg)(9.8m/s^2)(0.29m)=2.85J

Next, the potential energy is equal to kinetic energy of the block and the bullet at the beginning of its motion:

U=\frac{1}{2}(M+m)v^2\\\\v=\sqrt{2\frac{U}{M+m}}=\sqrt{2\frac{2.85J}{1kg+6*10^{-3}kg}}=2.38\frac{m}{s}

Next, you use the momentum conservation law, in order to calculate the speed of the bullet before the impact:

Mv_1+mv_2=(M+m)v    (2)

v1: initial velocity of the wood block = 0m/s

v2: initial speed of the bullet

v: speed of bullet and block = 2.38m/s

You solve the equation (2) for v2:

M(0)+mv_2=(M+m)v    

v_2=\frac{M+m}{m}v=\frac{1kg+6*10^{-3}kg}{6*10^{-3}kg}(2.38m/s)\\\\v_2=399.04\frac{m}{s}

The speed of the bullet before the impact with the wood block is 399.04 m/s

(b) The impulse is gibe by the change in the velocity of the block, multiplied by the mass of the block:

I=M\Delta v=M(v-v_1)=(1kg)(2.38m/s-0m/s)=2.38kg\frac{m}{s}

The impulse is 2.38 kgm/s

(c) The force on the cord after the impact is equal to the centripetal force over the block and bullet. That is:

T=F_c=(M+m)\frac{v^2}{l}=(1.006kg)\frac{(2.38m/s)^2}{2.2m}=2.59N    

The force on the cord after the impact is 2.59N

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3 years ago
Some machines do not multiply the force that is applied to them
Katyanochek1 [597]

Answer:

Machine Efficiency

Explanation:

Efficiency is the percent of work put into a machine by the user (input work) that becomes work done by the machine (output work). The output work is always less than the input work because some of the input work is used to overcome friction. Therefore, efficiency is always less than 100 percent

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A student observes a shadow move across the sun during a solar eclipse. Which list represents the position of Earth, the sun, an
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It goes sun moon earth the moon is blocking us from seeing the sun.
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