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Tresset [83]
3 years ago
6

A golf club with a mass of .85 kg hits a golf ball with a mass of .075 kg. The golf ball was initially at rest. After being hit,

it is moving at 35 m/s. The golf club and the golf ball were in contact for .015 s. Calculate the following: (4 points each part)
a) The acceleration of the golf ball


b) The force exerted on the golf ball by the club


c) The force exerted on the golf club by the ball


d) The acceleration of the golf club
Physics
1 answer:
Mariulka [41]3 years ago
8 0

Answer:

- أكثر أنواع التربة خصوبة التربة

- الحمراء .

.

ج- السوداء

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The info below shows three kettles with their powers and the time they take to boil 500cm3 of water. If electricity costs 9p per
dimaraw [331]

The cost of boiling 500 cm³ of water using the 3 KW Kettle is 1.35p

<h3>What is power? </h3>

This is defined as the rate in which energy is consumed. Electrical power is expressed mathematically as:

Power (P) = Energy (E) / time (t)

P = E / t

<h3>How to determine the energy</h3>
  • Power (P) = 3 KW
  • Time (t) = 3 mins = 3 / 60 = 0.05 h
  • Energy (E) =?

E = Pt

E = 3 × 0.05

E = 0.15 KWh

<h3>How to determine the cost</h3>
  • Energy (E) = 0.15 KWh
  • Cost per unit = 9p
  • Cost =?

Cost = Energy × cost per unit

Cost = 0.15 × 9

Cost = 1.35p

Learn more about electrical power:

brainly.com/question/64224

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8 0
1 year ago
A 2.0-kg laptop sits on the horizontal surface of the seat of a car moving at 8.0 m/s. The driver starts slowing down to stop. F
ivanzaharov [21]

Answer: 32.65\ m

Explanation:

Given

mass of laptop m=2 kg

The velocity of car u=8 m/s

The coefficient of static friction is \mu_s=0.4

The coefficient of kinetic friction is \mu_k=0.2

As the car is moving, so the coefficient of kinetic friction comes into play

deceleration offered by friction \mu_kg=0.2\times 9.8\ m/s^2

Using the equation of motion v^2-u^2=2as\\

insert the values

0^2-8^2=2(-0.2\times 9.8)s\\\\s=\dfrac{64}{1.96}\\\\s=32.65\ m

4 0
3 years ago
A solid disk of radius 5.50 cm and mass 1.25 kg , which is rolling at a speed of 1.50 m/s , begins rolling without slipping up a
shtirl [24]

Answer:

Explanation:

Deceleration of solid disk = g sin10/1 + k²/r² = g sin 10 / 1 + 1/2 = g sin 10 x 2/3

[ k is radius of gyration of disk which is equal to( 1/√2)x r ]

deceleration a = -1.1345 m/s²

v = u - at , t = u / a = 1.5 / 1.1345 = 1.322 s.

5 0
3 years ago
Consider a spherical volume of space that is large enough to be considered homogeneous. Also consider a particle on the surface
LenKa [72]

Answer:

Option A applies.

A.  Greater than its escape speed from the mass within the volume

Explanation:

Here it is mentioned that the spherical volume is large enough for the space to be considered as homogeneous. Also, the pressure within the volume is negligible, so that will not result into the re collapse of the Universe. Now as per our knowing, Hubble's Law relates the average speed of the particle to the distance R between the Earth and the particle. So, if the particle's speed is greater than it's escape speed from the mass within the volume, then the Universe is bound to re collapse back again. Option A applies.

3 0
3 years ago
A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t
vladimir1956 [14]

a) \omega=\frac{-mvR}{I+MR^2}

b) v=\frac{-mvR^2}{I+MR^2}

Explanation:

a)

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

L_1=0

Later, after the girl throws the rock, the angular momentum will be:

L_2=(I_M+I_g)\omega +L_r

where:

I is the moment of inertia of the merry-go-round

I_g=MR^2 is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

\omega is the angular speed of the merry-go-round and the girl

L_r=mvR is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

L_1=L_2

So we find:

0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

b)

The linear speed of a body in rotational motion is given by

v=\omega r

where

\omega is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

\omega=\frac{-mvR}{I+MR^2} is the angular speed

r=R is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

v=\omega R=\frac{-mvR^2}{I+MR^2}

5 0
2 years ago
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