A sports car accelerates from 0 to 25 meters per second in 4 seconds. What is its acceleration?

A rifle fires a bullet at a target. The speed of the bullet is 600m/s. The target is located 400m away. How long does it take fo

Alex

0.67s

Explanation:

Given parameters:

Speed of bullet = 600m/s

Distance of target = 400m

Unknown:

Time taken for bullet to reach target = ?

Solution:

Speed is a physical quantity that expresses the rate of change of distance with time;

Speed = $\frac{distance}{time taken}$

Since time is unknown, we make it the subject of the expression;

time = $\frac{distance }{speed}$ = $\frac{400}{600}$

time = 0.67s

Learn more:

Speed brainly.com/question/10048445

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5 0

3 years ago

Suppose the maximum power delivered by a car's engine results in a force of 16000 N on the car by the road. In the absence of an

joja [24]

Answer:

Approximately $9.7\; \rm m \cdot s^{-2}$ .

Explanation:

Assuming that there is no other force on this vehicle, the $16000\; \rm N$ force from the road would be the only force on this vehicle. The net force would then be equal to this $16000\; \rm N\!$ force. The size of the net force would be $16000\; \rm N\!\!$ .

Let $m$ denote the mass of this vehicle and let $\Sigma F$ denote the net force on this vehicle.

By Newton's Second Law of motion, the acceleration of this vehicle would be proportional to the net force on this vehicle. In other words, the acceleration of this vehicle, $a$ , would be:

$\begin{aligned}a &= \frac{\Sigma F}{m}\end{aligned}$ .

For this vehicle, $\Sigma F = 16000\; \rm N$ whereas $m = 1650\; \rm kg$ . The acceleration of this vehicle would be:

$\begin{aligned}a &= \frac{16000\; \rm N}{1650\; \rm kg} \\ &= \frac{16000\; \rm kg \cdot m\cdot s^{-2}}{1650\; \rm kg}\\ &\approx 9.7 \; \rm m \cdot s^{-2}\end{aligned}$ .

8 0

3 years ago

A skier leaves the horizontal end of a ramp with a velocity of 25.0 m/s and lands 70.0 m from the base of the ramp. How high is

Valentin [98]

Answer:

Explanation:

<u>Horizontal Motion </u>

When an object is thrown horizontally with an initial speed v and from a height h, it follows a curved path ruled by gravity.

The maximum horizontal distance traveled by the object can be calculated as follows:

$\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}$

If the maximum horizontal distance is known, we can solve the above equation for h:

$\displaystyle h=\frac {d^2g}{2v^2}$

The skier initiates the horizontal motion at v=25 m/s and lands at a distance d=70 m from the base of the ramp. The height is now calculated:

$\displaystyle h=\frac {70^2\cdot 9.8}{2\cdot 25^2}$

$\displaystyle h=\frac {4900\cdot 9.8}{2\cdot 625}$

h= 38.416 m

The end of the ramp is 38.416 m high

8 0

3 years ago

Arrange the stars based on their temperature. Begin with the coolest star, and end with the hottest star.

Slav-nsk [51]

Answer:

Uranus, Pluto, Neptune, Saturn , Jupiter, mars, Venus ,mercury and sun

4 0

2 years ago

Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the

Doss [256]

Answer:

$v_{su} = 19.44 m/s$

Explanation:

$m_{su}=5.68x10^{29}kg\\m_{sa}=5.68x10^{26}kg$

$T=9.29x10^8\\r_{o}=1.43x10^{12}$

If the sun considered as x=0 on the axis to put the center of the mass as a:

$m_{su}*r_{o}=(m_{sa}+m_{su})*r_{1}$

solve to r1

$r_1=\frac{m_{sa}*r_{o}}{m_{sa}+m_{su}}=\frac{5.68x10^{26}*1.43x10^{12}}{5.68x10^{26}+5.68x10^{26}}$

$r_1=1.428x10^9m$

Now convert to coordinates centered on the center of mass. call the new coordinates x' and y' (we won't need y'). Now since in the sun centered coordinates the angular momentum was

$L = \frac{m_{sa}*2*pi*r_1^2}{T}$