All categories

2 years ago

Calculate net force 532N 215N

Physics

1 answer:

lorasvet [3.4K]2 years ago

8 0

Answer:

FN is the forces acting on a body. When the body is at rest, the net force formula is given by, FNet = Fa + Fg.

Im in 7th and thats all I know so I hope it's enough

You might be interested in

Sound waves in air are a series of

GaryK [48]

Sound waves in air are a series of periodic disturbances, periodic condensations and rarefactions, and high- and low-pressure regions. It is all of the above. The answer is letter D.

5 0

2 years ago

A ray of light traveling in air strikes the surface of a liquid. if the angle of incidence is 29.7◦ and the angle of refraction

lana66690 [7]

When light moves from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle above which there is no refracted ray, and it is given by:
$\theta_c = \arcsin ( \frac{n_r}{n_i} )$ (2)
where $n_r$ is the refractive index of the second medium and $n_i$ is the refractive index of the first medium.

We can find the ratio $n_r / n_i$ by using Snell's law:
$n_i \sin \theta_i = n_r \sin \theta_r$ (1)
where
$\theta_i$ is the angle of incidence
$\theta_r$ is the angle of refraction

By using the data of the problem and re-arranging (1), we find
$\frac{n_r}{n_i} = \frac{\sin \theta_i}{\sin \theta_r} = \frac{\sin 16.3^{\circ}}{\sin 29.7^{\circ}} =0.566$

and if we use eq.(2) we can now find the value of the critical angle:
$\theta_c = \arcsin ( \frac{n_r}{n_i} ) = \arcsin (0.566) = 34.5^{\circ}$

3 0

3 years ago

1 lb equals how many grams

shusha [124]

Answer:

1 lb. is 453.592 grams

Explanation:

1lb is 453.592 grams

8 0

3 years ago

Read 2 more answers

Two workers are sliding 350 kgkg crate across the floor. One worker pushes forward on the crate with a force of 390 NN while the

svetoff [14.1K]

Answer:

$\mu_k=0.18$

Explanation:

First, we write the equations of motion for each axis. Since the crate is sliding with constant speed, its acceleration is zero. Then, we have:

$x: T+F-f_k=0\\\\y:N-mg=0$

Where T is the tension in the rope, F is the force exerted by the first worker, f_k is the frictional force, N is the normal force and mg is the weight of the crate.

Since $f_k=\mu_k N$ and $N=mg$ , we can rewrite the first equation as:

$T+F-\mu_k mg=0$

Now, we solve for $\mu_k$ and calculate it:

$\mu_k=\frac{T+F}{mg}\\ \\\mu_k =\frac{220N+390N}{(350kg)(9.8m/s^{2})} =0.18$

This means that the crate's coefficient of kinetic friction on the floor is 0.18.

6 0

3 years ago

On a straight road (taken to be in the x direction) you drive for an hour at 60 km per hour, then quickly speed up to 120 km per

Luda [366]

Answer:

The average velocity is 180 km/hr

Explanation:

Given;

initial velocity, u = 60 km per hour

final velocity, v = 120 km per hour

initial time = 1 hour

final time = 2 hour

Initial position = 60 km/h x 1 hour = 60 km

final position = 120 km/h x 2 hour = 240 km

The average velocity is given by;

$V_{avg} = \frac{Final \ position\ - \ Initial \ position}{final \ time\ - \ initial \ time}\\\\V_{avg} = \frac{240km \ - \ 60km}{2hr\ - \ 1hr} \\\\V_{avg} = \frac{180 \ km}{1hr} \\\\V_{avg}= 180 \ km/hr$

Therefore, the average velocity is 180 km/hr

3 0

3 years ago