All categories

2 years ago

50 N

Physics

1 answer:

Mariana [72]2 years ago

5 0

The work is done by the force pushing the mower is determined as 500 J.

<h3>Work done by the force</h3>

The work done by the force is calculated as follows;

W = Fsinθ x d

where;

Fsinθ = Fx is the applied force parallel to the ground
d is the distance
θ is the angle of inclination

W = Fx(d)

W = 25 x 20

W = 500 J

Thus, the work is done by the force pushing the mower is determined as 500 J.

Learn more about work done here: brainly.com/question/8119756

#SPJ1

You might be interested in

A boat is drifting with a speed of 5.0\,\dfrac{\text m}{\text s}5.0 s m 5, point, 0, space, start fraction, m, divided by, s,

zepelin [54]

Answer:

$d=42m(-\hat{i})$

Explanation:

For this problem, we need to apply the formulas of constant accelerated motion.

To obtain the boat displacement we need to calculate the displacement because of the river flow and the displacement done because of the boat motor.

for the river:

$d_r=v*t\\d_r=5m/s*6s\\d_r=30m(\hat{i})$

for the boat:

$x=\frac{1}{2}*a*t^2\\\\x=\frac{1}{2}*4.0m/s*(6s)^2\\\\\\x=72m(-\hat{i})$

So the final displacement is given by:

$d=dr+x\\d=30m-72m\\d=42m(-\hat{i})$

8 0

4 years ago

Read 2 more answers

What mass of a material with density rho is required to make a hollow spherical shell having inner radius r1 and outer radius r2

Margarita [4]

Answer:

$m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3 )$

Explanation:

We have to make a hollow sphere of inner radius $r_1$ and outer radius $r_2$ .

Then the mass of the material required to make such a sphere would be calculated as:

Total volume of the spherical shell:

$V_t=\frac{4}{3} \pi.r_2^3$

And the volume of the hollow space in the sphere:

$V_h=\frac{4}{3} \pi.r_1^3$

Therefore the net volume of material required to make the sphere:

$V=V_t-V_h$

$V=\frac{4}{3} \pi(r_2^3-r_1^3)$

Now let the density of the of the material be $\rho$ .

<u>Then the mass of the material used is:</u>

$m=\rho.V$

$m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3 )$

4 0

4 years ago

Bob and John are pulling in different directions. If Bob is pulling to the right with a force of 10N, and John is pulling to the

kykrilka [37]

Answer:

A) -2N

B) Left

C) -0.5

Explanation:

A) -12 + 10

B) More force is acted on in that direction

C) Net force/Mass (-2/4)

5 0

3 years ago

If the distance between slits on a diffraction grating is 0.50 mm and one of the angles of diffraction is 0.25°, how large is th

Trava [24]

Answer:

- path differnce = 2.18*10^-6

- 1538 lines

Explanation:

- The path difference for the waves that produce the pattern of diffraction, is given by the following formula:

$path\ difference\ = dsin\theta$ (1)

d: separation between slits = 0.50mm = 0.50*10^-3 m

θ: angle of a diffraction = 0.25°

Then, the path difference is:

$path\ difference\ =(0.50*10^{-3}m)sin(0.25\°)=2.18*10^{-6}m$

- The maximum number of bright lines are calculated by using the following formula:

$m\lambda = dsin\theta$ (2)

m: order of the bright

λ: wavelength = 650nm

The maximum bright is calculated for an angle of 90°:

$m=\frac{(0.50*10^{-3}m)sin90\°}{650*10^{-9}m} \approx 769$

The maxium number of bright lines are twice the previous result, that is, 1538 lines

8 0

3 years ago

Read 2 more answers

Sylvia and Jadon now want to work a problem. Imagine a puck of mass 0.5 kg moving in a circular simulation. Suppose that the ten

leva [86]

Answer:

Tangential speed, v = 2.64 m/s

Explanation:

Given that,

Mass of the puck, m = 0.5 kg

Tension acting in the string, T = 3.5 N

Radius of the circular path, r = 1 m

To find,

The tangential speed of the puck.

Solution,

The centripetal force acting in the string is balanced by the tangential speed of the puck. The expression for the centripetal force is given by :

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

$v=\sqrt{\dfrac{Fr}{m}}$

$v=\sqrt{\dfrac{3.5\ N\times 1\ m}{0.5\ kg}}$

v = 2.64 m/s

Therefore, the tangential speed of the puck is 2.64 m/s.

3 0

3 years ago