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4 years ago

a rock with mass of 5kg is carried up a small hill 10 meters high. how much work had to be done in carrying the rock up hill

Physics

1 answer:

aliya0001 [1]4 years ago

6 0

Answer:

490Nm

Explanation:

Given the following :

Mass of rock = 5kg

Height or distance = 10 meters

A rock with mass of 5kg is carried up a small hill 10 meters high. how much work had to be done in carrying the rock up hill?

Workdone is product of force and distance.

Workdone = force * distance

Recall :

Force = mass * acceleration due to gravity (g)

g = 9.8m/s^2

Force = 5 * 9.8 = 49 N

Therefore,

Workdone = 49 × 10 = 490Nm

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3 years ago

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Margy is trying to improve her cardio endurance by performing an exercise in which she alternates walking and running 100.0 m ea

Kipish [7]

Answer:

6.5 m/s

Explanation:

We are given that

Distance, s=100 m

Initial speed, u=1.4 m/s

Acceleration, $a=0.20 m/s^2$

We have to find the final velocity at the end of the 100.0 m.

We know that

$v^2-u^2=2as$

Using the formula

$v^2-(1.4)^2=2\times 0.20\times 100$

$v^2-1.96=40$

$v^2=40+1.96$

$v^2=41.96$

$v=\sqrt{41.96}$

$v=6.5 m/s$

Hence, her final velocity at the end of the 100.0 m=6.5 m/s

5 0

3 years ago

A soccer player extends her lower leg in a kicking motion by exerting a force with the muscle above the knee in the front of her

lisabon 2012 [21]

Answer:

10.09 N

Explanation:

Analogously to Newton's second law, torque can be defined as:

$\tau=I\alpha$

Here, I is the moment of inertia and $\alpha$ is the angular acceleration. We have:

$\tau=(0.65kg*m^2)(29.5\frac{rad}{s^2})\\\tau=19.18N*m$

Torque is the vector product of the position vector of the point at which the force is applied by the force vector:

$\vec{\tau}=\vec{r}\times \vec{F}$

Since the effective lever arm is perpendicular to the force, the angle between them is $90^\circ$ . The magnitud of this vector product is defined as:

$\tau=rFsen\theta$ .

Solving for F and replacing the known values:

$F=\frac{\tau}{rsen\theta}\\F=\frac{19.18N*m}{1.9m(sen90^\circ)}\\F=10.09N$

8 0

3 years ago

An earthquake emits both S-waves and P-waves which travel at different speeds through the Earth. A P-wave travels at 9 000 m/s a

Cerrena [4.2K]

Assuming constant speeds, the P-wave covers a distance d in time t such that

9000 m/s = d/(60 t)

while the S-wave covers the same distance after 1 more minute so that

5000 m/s = d/(60(t + 1))

Now,

d = 540,000 t

d = 300,000(t + 1) = 300,000 t + 300,000

Solve for t in the first equation and substitute it into the second equation, then solve for d :

t = d/540,000

d = 300,000/540,000 d + 300,000

4/9 d = 300,000

d = 675,000

So the earthquake center is 675,000 m away from the seismic station.

6 0

3 years ago

The first-order decomposition of cyclopropane has a rate constant of 6.7 × 10–4 s–1. If the initial concentration of cyclopropan

Ket [755]

Answer:

.864 M

Explanation:

For first order decomposition,

rate constant k = 1/t x ln a / (a - x )

given , a = 1.33 M , t = 644 s , k = 6.7 x 10⁻⁴ , a - x = ? = b( let )

6.7 x 10 ⁻⁴ = 1/644 x ln 1.33/b

ln 1.33/b = 6.7 x 10⁻⁴ x 644 = .4315

1.33 / b = e⁰ ⁴³¹⁵ = 1.5395

b = 1.33 / 1.5395 = .864 M.

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4 years ago