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3 years ago

10

An object traveling 200 feet per second slows to 50 feet per second in 5 seconds. Calculate the acceleration of the object. Use

the following formula:
Variable
Equation
Solve

1 answer:

shtirl [24]3 years ago

3 0

Answer:

-30m/s

Explanation:

Given:

Initial velocity of object = 200 feet/second

Final velocity of object = 50 feet/second

Time of travel = 5 seconds

To calculate acceleration of the object we will find the rate of change of velocity with respect to time.

So, acceleration "a" is given by:

$a=\frac{v_f-v_i}{t}$

where vf represents final velocity, vi represents initial velocity and is time of travel.

Plugging in values to evaluate acceleration.

$a=\frac{50-200}{5}$

$a=\frac{-150}{5}$

$a= -30m/s$

The acceleration of the object is -30m/s

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wo kids are on a seesaw. The one on the left has a mass of 75 kg and is sitting 1.5 m from the pivot point. The one on the right

Anastaziya [24]

Answer:

4.5 Nm (Anticlockwise)

Explanation:

Let the 75 kg kid is sitting at the left end and the 60 kg kid is sitting on the right end.

Anticlockwise Torque = 75 x 1.5 = 112.5 Nm

clockwise Torque = 60 x 1.8 = 108 Nm

Net torque = Anticlockwise torque - clockwise torque

Net Torque = 112.5 - 108 = 4.5 Nm (Anticlockwise)

5 0

3 years ago

Three particles lie in the xy plane. Particle 1 has mass m1 = 6.7 kg and lies on the x-axis at x1 = 4.2 m, y1 = 0. Particle 2 ha

krek1111 [17]

Answer:

$F=18.58\times 10^{-11}\ N$

$\theta=30.276^{\circ}$

Explanation:

Given:

mass of first particle, $m_1=6.7\ kg$

mass of second particle, $m_2=5.1\ kg$

mass of third particle, $m_3=3.7\ kg$

coordinate position of first particle in meters, $(x_1,y_1)\equiv(4.2,0)$

coordinate position of second particle in meters, $(x_2,y_2)\equiv(0,2.8)$

coordinate position of third particle in meters, $(x_3,y_3)\equiv(0,0)$

<u>Now, gravitational force on particle 3 due to particle 1:</u>

$F_{31}=G\frac{m_1.m_3}{r_{31}^2}$

$F_{31}=6.67\times 10^{-11} \times \frac{6.7\times 3.7}{4.2^2}$

$F_{31}=9.37\times 10^{-11}\ N$

towards positive Y axis.

<u>gravitational force on particle 3 due to particle 2:</u>

$F_{32}=G\frac{m_2.m_3}{r_{21}^2}$

$F_{32}=6.67\times 10^{-11} \times \frac{5.1\times 3.7}{2.8^2}$

$F_{32}=16.05\times 10^{-11}\ N$

towards positive X axis.

<u>Now the net force</u>

$F=\sqrt{F_{31}\ ^2+F_{32}\ ^2}$

$F=\sqrt{(10^{-11})^2(9.37^2+16.05^2)}$

$F=18.58\times 10^{-11}\ N$

<em>For angle in counterclockwise direction from the +x-axis</em>

$tan\theta=\frac{9.37\times 10^{-11}}{16.05\times 10^{-11}}$

$\theta=30.276^{\circ}$

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3 years ago

Is it possible to be moving but not be in motion?

GREYUIT [131]

No. Motion is the thing that when you're moving, you're in it.

But it IS possible for one person to say you're moving and another person to say you're not moving, both at the same time, and both of them are correct !

4 0

3 years ago

Read 2 more answers

A bicyclist starting at rest produces a constant angular acceleration of 1.30 rad/s2 for wheels that are 35.5 cm in radius.

Debora [2.8K]

Answer:

a) 0.462 m/s^2

b) 31.5 rad/s

c) 381 rad

d) 135m

Explanation:

the linear acceleration is given by:

$a=\alpha *r\\a=1.30rad/s^2*(35.5*10^{-2}m)\\a=0.462m/s^2$

the angular speed is given by:

$\omega=\frac{v}{r}\\\\\omega=\frac{11.2m/s}{35.5*10^{-2}m}\\\\\omega=31.5rad/s$

to calculate how many radians have the wheel turned we need the apply the following formula:

$\theta=\frac{1}{2}\alpha*t^2\\\\t=\frac{\omega}{\alpha}\\\\t=\frac{31.5rad/s}{1.30rad/s^2}\\\\t=24.2s\\\\\theta=\frac{1}{2}*1.30rad/s^2*(24.2s)^2\\\\\theta=381rad$

the distance is given by:

$d=\theta*r$

$d=381rad*(35.5*10^{-2}m)\\d=135m$

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4 years ago

using newtons law a force of 250N is applied to an object that accelerates at a rate of 5M/s2 what is the mass of the object?

AURORKA [14]

Answer:

50 kg

Explanation:

F = ma

250 N = m (5 m/s²)

m = 50 kg

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3 years ago