All categories

2 years ago

A man hits a golf ball (0.2 kg) which accelerates at a rate of 20 m/s2. What amount of force acted on the ball?

1 answer:

4 0

Newton taught us that Force = (mass) x (acceleration)

Force = (0.2) x (20) = 4 newtons .

Something to think about: The ball can only accelerate while the club-face
is in contact with it. Once the ball leaves the club, it can't accelerate any more,
because the force against it is gone.

You might be interested in

Se lanza una pelota de béisbol desde la azotea de un edificio de 25 m de altura con velocidad inicial de magnitud 10 m/s y dirig

MissTica

Answer:

v_f = 24.3 m / s

Explanation:

A) In this exercise there is no friction so energy is conserved.

Starting point. On the roof of the building

Em₀ = K + U = ½ m v₀² + m g y₀

Final point. On the floor

Em_f = K = ½ m v_f²

Emo = Em_g

½ m v₀² + m g y₀ = ½ m v_f²

v_f² = v₀² + 2 g y₀

let's calculate

v_f = √(10² + 2 9.8 25)

v_f = 24.3 m / s

6 0

2 years ago

In handball, who is the only player allowed in the goalie area?

lidiya [134]

Answer:

Only the goalie is allowed inside the goal crease. The only exception when another player is allowed in the goal area is when they take off from outside the goal area, and shoots or passes the ball before landing. To avoid interference with other players, the player must then exit the goal area as soon as possible.

Explanation:

4 0

2 years ago

Helpp .....meeee pleaseee

Lady bird [3.3K]

Answer:

hydrogen

helium

oxygen

Explanation:

join this grop to get instant answer

it's very helpful

5 0

2 years ago

-. A 2kg cart moving to the right at 5m/s collides with an 8kg cart at rest. As a

bulgar [2K]

Answer:

The velocity of the carts after the event is 1 m/s

Explanation:

Law Of Conservation Of Linear Momentum

The total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is

P=mv.

If we have a system of bodies, then the total momentum is the sum of the individual momentums:

$P=m_1v_1+m_2v_2+...+m_nv_n$

If a collision occurs and the velocities change to v', the final momentum is:

$P'=m_1v'_1+m_2v'_2+...+m_nv'_n$

Since the total momentum is conserved, then:

P = P'

In a system of two masses, the equation simplifies to:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

If both masses stick together after the collision at a common speed v', then:

$m_1v_1+m_2v_2=(m_1+m_2)v'$

The common velocity after this situation is:

$\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

The m1=2 kg cart is moving to the right at v1=5 m/s. It collides with an m2= 8 kg cart at rest (v2=0). Knowing they stick together after the collision, the common speed is:

$\displaystyle v'=\frac{2*5+8*0}{2+8}=\frac{10}{10}=1$

The velocity of the carts after the event is 1 m/s

3 0

2 years ago

A block of ice at 0°C is added to a 150g aluminum calorimeter cup which holds 210 g of water at 12°C. If all but 2.0 g of ice me

Over [174]

Answer:

original mass of the block of ice is 38.34 gram

Explanation:

Given data

cup mass = 150 g

ice temperature = 0°C

water mass = 210 g

water temperature = 12°C

ice melt = 2 gram

to find out

solution

we know here

specific heat of aluminum is c = 0.900 joule/gram °C

Specific heat of water C = 4.186 joule/gram °C

so here temperature difference is dt = 12- 0 = 12°C

so here heat lost by water and cup are given by

heat lost = cup mass × c × dt + water mass × C × dt

heat lost = 150 × 0.900 × 12 + 210 × 4.186 × 12

heat lost = 12168.72 J

mass of ice melt here = heat lost / latent heat of fusion

here we know latent heat of fusion = 334.88 joule/gram

mass of ice melt = 12168.72 / 334.88

mass of ice melt is 36.337554 gram

so mass of ice is here = mass of ice melt + ice melt

mass of ice = 36.337554 + 2

mass of ice = 38.337554 gram

so original mass of the block of ice is 38.34 gram

7 0

2 years ago