Ask question

Ask question

All categories

3 years ago

6

Two horses, Thunder and Misty, are accelerating a wagon 1.3 m/s2. The force of friction is 75 N. Thunder is pulling with a force

of 1,000 N, while Misty is pulling with a force of 800 N. The force of gravity is 14,700 N and the normal force is 14,700 N. What is the mass of the wagon? Round the answer to the nearest whole number.

1 answer:

just olya [345]3 years ago

3 0

Answer:

1327 kg

Explanation:

So the net force exerted on the wagon would be the sum of forces from 2 horses subtracted by friction force

$F = 1000 + 800 - 75 = 1725 N$

This force results in an acceleration of a = 1.3 m/s2. We can use Newton's 2nd law to calculate the mass of the wagon

$F = ma$

$m = F / a = 1725 / 1.3 \approx 1327 kg$

You might be interested in

What does the word "atom"

Oliga [24]

Answer:

in greek atom means uncuttable

7 0

2 years ago

Read 2 more answers

Match each influence to its correct category

Pie

Answer:

healthful I think! Hope that helped

6 0

2 years ago

Read 2 more answers

Light is incident on the left face of an isosceles prism; with an apex angle of 49o, such that the light exiting the right face

Sunny_sXe [5.5K]

Answer:

$\mu = 1.645$

Explanation:

By Snell's law we know at the left surface

$\theta_i = 19^o$

$\theta_r = ?$

$\mu_1 = 1$

$\mu_2 = \mu$

now we have

$1 sin19 = \mu sin\theta_r$

$0.33 = \mu sin\theta_r$

now on the other surface we know that

angle of incidence = $\theta_r'$

$\theta_e = 90$

so again we have

$\mu sin\theta_r' = 1 sin90$

so we have

$\theta_r = sin^{-1}\frac{0.33}{\mu}$

$\theta_r' = sin^{-1}\frac{1}{\mu}$

also we know that

$\theta_r + \theta_r' = 49$

$sin^{-1}\frac{0.33}{\mu} + sin^{-1}\frac{1}{\mu} = 49$

By solving above equation we have

$\mu = 1.645$

3 0

3 years ago

Read 2 more answers

Why are male genitals outside the body?

BigorU [14]

Because if they were in the inside, then they would not be able to sexually reproduct and could not have offspring.

7 0

3 years ago

How much mass should be attached to a vertical ideal spring having a spring constant (force constant) of 39.5 n/m so that it wil

mrs_skeptik [129]

The frequency of a simple harmonic oscillator such as a spring-mass system is given by
$f= \frac{1}{2 \pi} \sqrt{ \frac{k}{m} }$
where
k is the spring constant
m is the mass attached to the spring.

Re-arranging the formula, we get:
$m= \frac{k}{4 \pi^2 f^2}$
and since we know the constant of the spring:
$k=39.5 N/m$
and the frequency of oscillation:
f=1.00 Hz
we can find the value of the mass attached to it:
$m= \frac{39.5 Hz}{4 \pi^2 (1.00 Hz)^2} = 1.00 kg$

7 0

3 years ago