All categories

3 years ago

If a force of 450N pushes a box 4 meters in 2 seconds, how much power was developed?

1 answer:

5 0

Work = (force) x (distance) = (450 N) x (4 m) = 1,800 joules

Power = (work) / (time) = (1,800 joules) / (2 sec) = 900 watts .

You might be interested in

If a car increases its velocity from zero to 60 m/s in 10 seconds, its acceleration is

andrezito [222]

We know that a=vf_vi/t equals equation "a" . Where a is the acceleration of the body , vf is the final velocity , vi is the initial velocity and t is equal to time . Since vi equals o m/s , vf equals to 60 m/s and t equals 10 s. Put in equation "a". a=60-0/10 =6m/s2

5 0

3 years ago

Sometimes i wish i wasnt alive....

pshichka [43]

Answer:

Swim, jogging, weight lifting are some activities to do. This will help the overall health of the heart and mind.

Explanation:

5 0

3 years ago

Read 2 more answers

A 58 kg skier is going down a 35 degree slope. The areaof each

maxonik [38]

To solve this problem we will use a free body diagram that allows us to determine the Normal Force.

In general, the normal force would be equivalent to

$N = mgcos\theta$

Since the skier is standing on two skis, his weight will be divide by two

$N' = \frac{mgcos\theta}{2}$

Pressure is given as the force applied in a given area, that is

$P = \frac{F}{A}$

Replacing F with N'

$P = \frac{N'}{A}$

$P = \frac{\frac{mgcos\theta}{2}}{A}$

Our values are given as,

$m = 58kg$

$g = 9.8m/s^2$

$\theta = 35\°$

$A = 0.3m^2$

Replacing we have that

$P = \frac{\frac{(58)(9.8)cos(35)}{2}}{0.3}$

$P = 776.01Pa$

Therefore the pressure exerted by each ski on the snow is 776.01Pa

6 0

3 years ago

What fraction of the total energy of a SHO is kinetic when the displacement is one third the amplitude

Bas_tet [7]

Answer:

The fraction of kinetic energy to the total energy is $\frac{K}{T}=\frac{8}{9}$ .

Explanation:

displacement is one third of the amplitude.

Let the amplitude is A.

x= A/3

The kinetic energy of the body executing SHM is

$K = 0.5 mw^2(A^2 - x^2)\\\\K = 0.5 m w^2 \left ( A^2 -\frac{A^2}{9} \right )\\\\K = 0.5 mw^2\times \frac{8A^2}{9}......(1)$

The total energy is

$T =0.5 mw^2A^2 ..... (2)$

Divide (1) by (2)

$\frac{K}{T}=\frac{8}{9}$

5 0

3 years ago

What is the minimum coefficient of friction needed for a frightened driver to take the same curve at 20.0 km/h?

Tanya [424]

Answer:

The minimum coefficient of friction is 0.22

Explanation:

Suppose If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the curve.

We need to calculate the ideal speed to take a 85 m radius curve banked at 15°.

Given that,

Radius = 85 m

Angle = 15°

Speed = 20 km/h

We need to calculate the ideal speed

Using formula of speed

$\tan\theta=\dfrac{v^2}{rg}$

$v=\sqrt{rg\tan\theta}$

Put the value into the formula

$v=\sqrt{85\times9.8\tan15}$

$v=14.9\ m/s$

We need to calculate the minimum coefficient of friction

Using formula for coefficient of friction

$v^2=\dfrac{rg(\sin\theta-\mu\cos\theta)}{\mu\sin\theta+\cos\theta}$

Put the value into the formula

$(5.55)^2=\dfrac{85\times9.8(\sin15-\mu\cos15)}{\mu\sin15+\cos15}$

$\dfrac{30.8025}{85\times9.8}=\dfrac{0.2588-\mu0.966}{\mu0.2588+0.966}$

$0.9754462\ mu-0.223541=0$

$\mu=\dfrac{0.223541}{0.9754462}$

$\mu=0.22$

Hence, The minimum coefficient of friction is 0.22

3 0

3 years ago