Ask question

Ask question

All categories

3 years ago

13

A pronghorn antelope has been observed to run with a top speed of 97 km/h. Suppose an antelope runs 1.5 km with an average speed

of 85 km/h, then runs 0.80 km with an average speed of 67 km/h

1 answer:

Angelina_Jolie [31]3 years ago

4 0

Answer:

10

Explanation:

You might be interested in

What is the impulse when a test car traveling at 15 m/s strikes a cement wall at a force of 1,900,000 N if the impact lasts for

svp [43]

Answer:

impulse = force x time

impulse = 1900000 x 0.005 = 9500Ns

8 0

3 years ago

A 54.0 kg ice skater is gliding along the ice, heading due north at 4.10 m/s . The ice has a small coefficient of static frictio

lesya [120]

Answer:

a. 2.668 m/s

b. 0.00494

Explanation:

The computation is shown below:

a. As we know that

$W = F\times d$

$KE = 0.5\times m\times v^2$

As the wind does not move the skater to the east little work is performed in this direction. All the work goes in the direction of the N-S. And located in that direction the component of the Force.

F = 3.70 cos 45 = 2.62 N

$W = F \times d = 2.62 N \times 100 m$

$W = 261.6 N\times m$

We know that

KE1 = Initial kinetic energy

KE2 = kinetic energy following 100 m

The energy following 100 meters equivalent to the initial kinetic energy less the energy lost to the work performed by the wind on the skater.

So, the equation is

KE2 = KE1 - W

$0.5 m\times v2^2 = 0.5 m\ v1^2 - W$

Now solve for v2

$v2 = \sqrt{v1^2 - {\frac{2W}{M}}}$

$= \sqrt{4.1 m/s)^2 - \frac{2 \times 261.6 N\times m}{54.0 kg}}$

= 2.668 m/s

b. Now the minimum value of Ug is

As we know that

Ff = force of friction

Us = coefficient of static friction

N = Normal force = weight of skater

So,

$Ff = Us\times N$

Now solve for Us

$= \frac{Ff}{N}$

$= \frac{3.70 N \times cos 45 }{54.0 kg \times 9.81 m/s^2}$

= 0.00494

4 0

3 years ago

describe how your data supports the statement the rate of movement of dna fragments is inversely proportional to the log of thei

wlad13 [49]

Answer: Is there supposed to be something elses here because this question is unclear.

Explanation:

5 0

3 years ago

A 50.0-kg crate is being pulled along a horizontal smooth surface. The pulling force is 10.0 n and is directed 20.0° above the h

steposvetlana [31]

Answer:

$0.188 m/s^2$

Explanation:

Assuming the crate does not lift above the ground and remains along the floor, then its acceleration will be in the horizontal direction. Therefore, we can use Newton's second law to find its acceleration:

$F_x = ma_x$

where

$F_x$ is the net force on the crate along the x-direction

m is the mass of the crate

$a_x$ is the acceleration

Here we have:

m = 50.0 kg

$F_x = F cos \theta = (10.0 N)(cos 20.0^{\circ})=9.4 N$ is the component of the pulling force along the horizontal direction

Solving for the acceleration,

$a_x = \frac{F_x}{m}=\frac{9.4}{50.0}=0.188 m/s^2$

6 0

3 years ago

What is newton's second law of motion?

KATRIN_1 [288]

<span>the acceleration of an object is depedent upon two variables

</span>

4 0

3 years ago

Read 2 more answers