The forces acting on a falling leaf are.

A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s

lisabon 2012 [21]

Answer:

Magnitude of the average force exerted on the wall by the ball is 800N

Explanation:

Given

Contact Time = t = 0.05 seconds

Mass (of ball) = 0.80kg

Initial Velocity = u = 25m/s

Final Velocity = 25m/s

Magnitude of the average force exerted on the wall by the ball is given by;

F = ma

Where m = 0.8kg

a = Average Acceleration

a = (u + v)/t

a = (25 + 25)/0.05

a = 50/0.05

a = 1000m/s²

Average Force = Mass * Average Acceleration

Average Force = 0.8kg * 1000m/s²

Average Force = 800kgm/s²

Average Force = 800N

Hence, the magnitude of the average force exerted on the wall by the ball is 800N

3 0

3 years ago

Derive an expression for the gravitational potential energy U(r) of the object-earth system as a function of the object's distan

Drupady [299]

Answer:

$U(r)=-\frac{Gm_Emr^2}{2R^3_E}$

Explanation:

We are given that

Gravitational force= $F_g=\frac{Gm_Emr}{R^3_E}$

r=0,U(0)=0

We know that

Gravitational potential energy= $-\int F_gdr$

$U(r)=-\int\frac{Gm_Emr}{R^3_E}dr$

$U(r)=-\frac{Gm_Em}{R^3_E}\times \frac{r^2}{2}+C$

Substitute r=0 ,U(0)=0

$0=0+C$

$C=0$

Substitute the value

$U(r)=-\frac{Gm_Emr^2}{2R^3_E}$

4 0

3 years ago

A wooden cube with the mass of 1kg is placed on a frictionless plane that makes an angle of 30° with the floor.

Anni [7]

Answer:

I dont know

Explanation:

8 0

2 years ago

a stone with a mass of 2.40 kg is moving with velocity (6.60î − 2.40ĵ) m/s. find the net work (in j) on the stone if its velocit

ch4aika [34]

By the work energy theorem, the total work done on the stone is given by its change in kinetic energy,

$W = \Delta K = \dfrac m2 ({v_2}^2 - {v_1}^2)$

We have

$\vec v_1 = (6.60\,\vec\imath - 2.40\,\vec\jmath)\dfrac{\rm m}{\rm s} \implies {v_1}^2 = \|\vec v_1\|^2 = 49.32 \dfrac{\rm m^2}{\rm s^2}$

$\vec v_2 = (8.00\,\vec\imath + 4.00\,\vec\jmath) \dfrac{\rm m}{\rm s} \implies {v_2}^2 = \|\vec v_2\|^2 = 80.0\dfrac{\mathrm m^2}{\mathrm s^2}$

Then the total work is

$W = \dfrac{2.40\,\rm kg}2 \left(80.0\dfrac{\rm m^2}{\rm s^2} - 49.32\dfrac{\rm m^2}{\rm s^2}\right) \approx \boxed{36.8\,\rm J}$

5 0

1 year ago

A sailboat runs before the wind with a constant speed of 2.8 m/s in a direction 52° north of west. How far (a) west and (b) nort

vodka [1.7K]

<h2>Displacement along west = 3612 m</h2><h2>Displacement along north = 4633.50 m</h2>

Explanation:

Let east be positive x axis and north be positive y axis

Velocity of boat = 2.8 m/s in a direction 52° north of west.

Velocity, v = -2.8 cos 52 i + 2.8 sin 52 j = -1.72 i + 2.21 j m/s

Time taken = 35 min = 35 x 60 = 2100 s

Displacement = Velocity x Time

Displacement = (-1.72 i + 2.21 j) x 2100

Displacement = -3612 i + 4633.50 j m

Displacement along west = 3612 m

Displacement along north = 4633.50 m

6 0

3 years ago