Ask question

Ask question

All categories

3 years ago

8

An ostrich can run at a speed of 43 mi/hr. How much ground can an ostrich cover if it runs at this speed for 15 minutes? (Hint:

15 minutes = 0.25 hours)

1 answer:

Simora [160]3 years ago

4 0

..... It would possibly she eenejjsjejeej 1.4

You might be interested in

If a ball is launched horizontally at 40 m/s

vaieri [72.5K]

Answer:

40m/s

Explanation:

The horizontal component of velocity remains constant because there are no external forces in that direction

By applying motion equations, V= U+ at

where ,

v - final velocity
u - initial velocity
a-acceleration
t - time

v = u +at

As no force act on the ball ( we neglect air resistance here) no acceleration is seen,

So v = u = 40m/s

4 0

3 years ago

A physics major is cooking breakfast when he notices that the frictional force between the steel spatula and the Teflon frying p

Semenov [28]

Answer:

$f_n=3.75N$

Explanation:

From the question we are told that:

Frictional force $F=0.150N$

Coefficient of kinetic friction $\mu=0.04$

Generally the equation for Normal for is mathematically given by

$f_n=\frac{F}{\mu}$

Therefore

$f_n=\frac{0.150}{0.04}$

$f_n=3.75N$

5 0

3 years ago

How do reflection and refraction of waves affect communication?

romanna [79]

Answer:

Reflection involves a change in direction of waves when they bounce off a barrier. Refraction of waves involves a change in the direction of waves as they pass from one medium to another.

5 0

3 years ago

Read 2 more answers

How much heat does it take to raise the temperature of 10.0 kg of water by 1.0 C?

fomenos

Answer:The specific heat capacity of water is 4,200 joules per kilogram per degree Celsius (J/kg°C). This means that it takes 4,200 J to raise the temperature of 1 kg of water by 1°C.

Explanation:

7 0

3 years ago

A 51.0 kg crate, starting from rest, is pulled across a floor with a constant horizontal force of 225 N. For the first 10.0 m th

shepuryov [24]

Answer:

The final speed of the crate is 12.07 m/s.

Explanation:

For the first 10.0 meters, the only force acting on the crate is 225 N, so we can calculate the acceleration as follows:

$F = ma$

$a = \frac{F}{m} = \frac{225 N}{51.0 kg} = 4.41 m/s^{2}$

Now, we can calculate the final speed of the crate at the end of 10.0 m:

$v_{f}^{2} = v_{0}^{2} + 2ad_{1}$

$v_{f} = \sqrt{0 + 2*4.41 m/s^{2}*10.0 m} = 9.39 m/s$

For the next 10.5 meters we have frictional force:

$F - F_{\mu} = ma$

$F - \mu mg = ma$

So, the acceleration is:

$a = \frac{F - \mu mg}{m} = \frac{225 N - 0.17*51.0 kg*9.81 m/s^{2}}{51.0 kg} = 2.74 m/s^{2}$

The final speed of the crate at the end of 10.0 m will be the initial speed of the following 10.5 meters, so:

$v_{f}^{2} = v_{0}^{2} + 2ad_{2}$

$v_{f} = \sqrt{(9.39 m/s)^{2} + 2*2.74 m/s^{2}*10.5 m} = 12.07 m/s$

Therefore, the final speed of the crate after being pulled these 20.5 meters is 12.07 m/s.

I hope it helps you!

7 0

3 years ago