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3 years ago

7

A car has an initial velocity of +30.0 m/s and undergoes an acceleration of -5.00 m/s squared for 5.00 seconds what is the displ

acement after the acceleration?

1 answer:

Leni [432]3 years ago

4 0

Ok so the formula is d=vi(t)+½at² and when you substitute it you should get 172.5meters

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Jan ran 4 miles north in 28 minutes. What was Jan's average velocity?

fenix001 [56]

Answer:

3.83 m/s

Explanation:

Given that,

Distance covered by Jan, d = 4 miles

1 mile = 1609.34 m

4 miles = 6437.38 m

Time, t = 28 minutes = 1680 s

Jan's average speed,

v = d/t

$v=\dfrac{6437.38\ \text{m}}{1680\ \text{s}}\\\\v=3.83\ \text{m/s}$

Hence, the average velocity of Jan is 3.83 m/s.

6 0

3 years ago

a particle of mass m sits at rest at x = 0. At time t = 0 a force given by F = Fe^(-t/T) is applied in the +x direction; F and T

wlad13 [49]

Explanation:

Given: $F=m\ddot{x}=Fe^{-\frac{t}{T}}$

Solving for $\ddot{x}$ :

$\ddot{x}=\frac{F}{m}e^{-\sqrt{\frac{F}{m} } t}$

where:

$T=\sqrt{\frac{m}{F}}$

Integrating to get $\dot{x}$ with initial conditions $\dot{x}(0)=0$ :

$\dot{x}=\sqrt{\frac{F}{m}}-\sqrt{\frac{F}{m}} e^{-\sqrt{\frac{F}{m}} t}$

Integrating to get x with initial conditions x(0) = 0:

$x=-1+\sqrt{\frac{F}{m}} t+e^{-\sqrt{\frac{F}{m}}t}$

When t=T:

$x=-1+\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}+e^{-\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}}=\frac{1}{e}$

$\dot{x}=\sqrt{\frac{F}{m}}-\sqrt{\frac{F}{m}} e^{-\sqrt{\frac{F}{m}}\sqrt{\frac{m}{F}}}=\sqrt{\frac{F}{m}}(1-\frac{1}{e})$

4 0

2 years ago

Consider an old-fashion bicycle with a small wheel of radius 0.17 m and a large wheel of radius 0.92 m. Suppose the rider starts

Gala2k [10]

Answer:

10259.6 m

Explanation:

We are given that

Radius of small wheel,r=0.17 m

Radius of large wheel,r'=0.92 m

Initial velocity,u=0

Time,t=2.7 minutes=162 s

1 min=60 s

Velocity,v=10m/s

Time,t'=13.7 minutes=822 s

Time,t''=4.1 minutes=246 s

$v=u+at$

Substitute the values

$10=0+162a=162a$

$a=\frac{10}{162}=0.0617m/s^2$

$s=ut+\frac{1}{2}at^2$

Substitute the values

$s=\frac{1}{2}(0.0617)(162)^2=809.6 m$

$s'=vt'=10\times 822=8220 m$

$a'=\frac{v}{t''}=\frac{10}{246}$

$s''=\frac{1}{2}a't''^2=\frac{1}{2}\times \frac{10}{246}(246)^2=1230 m$

Total distance traveled by rider=s+s'+s''=809.6+8220+1230=10259.6 m

6 0

3 years ago

Read 2 more answers

(8c8p49) A 115g Frisbee is thrown from a point 1.00 m above the ground with a speed of 12.00 m/s. When it has reached a height o

IgorLugansk [536]

Answer:

The work done on the Frisbee is 1.36 J.

Explanation:

Given that,

Mass of Frisbee, m = 115 g = 0.115 kg

Initial speed of Frisbee, u = 12 m/s at a point 1 m above the ground

Final speed of Frisbee , v = 10.9674 m/s when it has reached a height of 2.00 m. Let W is the work done on the Frisbee by its weight. According to work energy theorem, the work done is equal to the change in its kinetic energy. So,

$W=\dfrac{1}{2}m(v^2-u^2)\\\\W=\dfrac{1}{2}\times0.115\times\left((10.9674)^{2}-(12)^{2})\right)\\\\W=-1.36\ J$

So, the work done on the Frisbee is 1.36 J. Hence, this is the required solution.

3 0

3 years ago

Read 2 more answers

3. An athlete makes a long jump and follows a projectile motion. Air resistance is negligible. Which one of the following statem

yulyashka [42]

Answer:

Option (b) is correct.

Explanation:

The motion under the influence of gravity is called projectile motion.

The acceleration due to gravity is constant through out the motion and it is always acting downwards.

When an athlete jumps and follow the projectile path, it always have the same horizontal velocity as there is no acceleration in the horizontal direction.

Also he has the vertical acceleration constant which is equal to the acceleration due to gravity and acts towards the center of earth.

Option (b) is correct.

6 0

3 years ago