All categories

2 years ago

A 6.5 x 104 W engine exerts a constant force on of 5.5 x 103 N on a car, the resulting velocity is?

Physics

1 answer:

jek_recluse [69]2 years ago

4 0

Answer:

12m/s

Explanation:

Given parameters:

Power = 6.5 x 10⁴W

Force = 5.5 x 10³N

Unknown:

The resulting velocity = ?

Solution:

The velocity of a body is related to force and power using the expression below;

Power = Force x velocity

Insert the parameters and solve for velocity

6.5 x 10⁴ = 5.5 x 10³ x velocity

velocity = $\frac{6.5 x 10^{4} }{5.5 x 10^{3} }$ = 12m/s

You might be interested in

As shown in the picture above, a sled of mass 67 kilograms is pulled

Scorpion4ik [409]

Explanation:

F net of sled = Tension force by rope - Kinetic friction between ground.

F normal of sled = mg = (67kg)(9.81kg/m^2) = 657.27N.

Kinetic friction = 0.18 (I cannot see the value) * Normal force of sled = 0.18 * 657.27N = 118.31N

So F net of sled = 800N - 118.31N = 681.69N.

(I cannot see what the question is asking for, please check on your own!)

3 0

3 years ago

WILL GIVE BRAINLIEST!!<br> What is a way to transfer charge in which an object becomes polarized?

Andreyy89

Answer:

I answered Number 4 (Solids and Elasticity)

Explanation:

solids and elasticity

4 0

2 years ago

Blood in a carotid artery carrying blood to the head is moving at 0.15 m/s when it reaches a section where plaque has narrowed t

sp2606 [1]

Answer:

26.9 Pa

Explanation:

We can answer this question by using the continuity equation, which states that the volume flow rate of a fluid in a pipe must be constant; mathematically:

$A_1 v_1 = A_2 v_2$ (1)

where

$A_1$ is the cross-sectional area of the 1st section of the pipe

$A_2$ is the cross-sectional area of the 2nd section of the pipe

$v_1$ is the velocity of the 1st section of the pipe

$v_2$ is the velocity of the 2nd section of the pipe

In this problem we have:

$v_1=0.15 m/s$ is the velocity of blood in the 1st section

The diameter of the 2nd section is 74% of that of the 1st section, so

$d_2=0.74d_1$

The cross-sectional area is proportional to the square of the diameter, so:

$A_2=(0.74)^2 A_1=0.548 A_1$

And solving eq.(1) for v2, we find the final velocity:

$v_2=\frac{A_1 v_1}{A_2}=\frac{A_1 (0.15)}{0.548 A_1}=0.274 m/s$

Now we can use Bernoulli's equation to find the pressure drop:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$

where

$\rho=1025 kg/m^3$ is the blood density

$p_1,p_2$ are the initial and final pressure

So the pressure drop is:

$p_1 - p_2 = \frac{1}{2}\rho (v_2^2-v_1^2)=\frac{1}{2}(1025)(0.274^2-0.15^2)=26.9 Pa$

8 0

2 years ago

What combination of variables will make an experiment the most reliable

Virty [35]

Answer:

Explanation:

A combination of two controlled variables will make an experiment the most reliable.

Variables are the values that we take under any circumstance while doing an experiment so that we can keep on changing and get new results at the end.

It is important to have them in pair so that two values can be kept on changing in terms of any constant condition. This will help to get better results in over all experiment data.

4 0

2 years ago

F = 2.0*10^20 N,

natka813 [3]

$F=\dfrac{Gm_1m_2}{r^2}$

With the given values of $F,G,m_1,r$ , we have

$2.0\times10^{20}\,\mathrm N=\dfrac{\left(6.67\times10^{-11}\,\frac{\mathrm{Nm}^2}{\mathrm{kg}^2}\right)\left(5.98\times10^{24}\,\mathrm{kg}\right)m_2}{\left(3.8\times10^8\,\mathrm m\right)^2}$