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sergey [27]
3 years ago
12

A baseball strikes the catcher's glove with a horizontal velocity of 40 m/s. The mass of the baseball is 0.15 kg. The displaceme

nt of the baseball due to the deformation of the catcher's glove and the movement of the catcher's hand is 8 cm from the instant it first makes contact with the glove until it stops. a. How much kinetic energy does the baseball possess just before it strikes the glove
Physics
1 answer:
choli [55]3 years ago
4 0

Answer:

the kinetic energy is 120 J

Explanation:

a. The computation of the kinetic energy before it strikes the glove is as follows:

As we know that

The Kinetic energy is

=  1 ÷ 2 × m × v^2

= 1 ÷ 2 × (0.15)(40^2)

= 120 J

Hence, the kinetic energy is 120 J

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A 1.4-kg cart is attached to a horizontal spring for which the spring constant is 50 N/m . The system is set in motion when the
marin [14]

Answer:

A)

0.395 m

B)

2.4 m/s

Explanation:

A)

m = mass of the cart = 1.4 kg

k = spring constant of the spring = 50 Nm⁻¹

x = initial position of spring from equilibrium position = 0.21 m

v_{i} = initial speed of the cart = 2.0 ms⁻¹

A = amplitude of the oscillation = ?

Using conservation of energy

Final spring energy = initial kinetic energy + initial spring energy

(0.5) kA^{2} = (0.5) m v_{i}^{2} + (0.5) k x_{i}^{2} \\kA^{2} = m v_{i}^{2} + k x_{i}^{2} \\(50) A^{2} = (1.4) (2.0)^{2} + (50) (0.21)^{2} \\A = 0.395 m

B)

m = mass of the cart = 1.4 kg

k = spring constant of the spring = 50 Nm⁻¹

A = amplitude of the oscillation = 0.395 m

v_{o} = maximum speed at the equilibrium position

Using conservation of energy

Kinetic energy at equilibrium position = maximum spring potential energy at extreme stretch of the spring

(0.5) m v_{o}^{2} = (0.5) kA^{2}\\m v_{o}^{2} = kA^{2}\\(1.4) v_{o}^{2} = (50) (0.395)^{2}\\v_{o} = 2.4 ms^{-1}

5 0
3 years ago
Read 2 more answers
The universe was 5 percent its current size when light left objects observed now at redshift of ______________
Nuetrik [128]

The redshift of distant galaxy are larger than those of closer galaxies, which indicates that the galaxy is receding at a faster rate.

  • The Universe was 5 percent its current size when light left objects now at redshift of <u>19</u>.

Reasons:

The size of the universe represented as a scale factor with relation to the redshift can be presented as follows;

\displaystyle \frac{a}{a_0}  =\mathbf{ \frac{1}{1 + z}}

Where;

a₀ = The current size of the Universe

a = The size of the early Universe = 5% of a

Therefore;

\displaystyle \frac{a}{a_0}  =5\% = 0.05=  \frac{1}{1 + z}

\displaystyle 0.05  = \frac{1}{1 + z}

0.05 + 0.05·z = 1

\displaystyle z = \mathbf{ \frac{1 - 0.05}{0.05} } = 19

  • The redshift is of the observed light is, z = <u>19</u>

Learn more here:

brainly.com/question/14459434

brainly.com/question/3654558

4 0
2 years ago
A 10.0-kg mass is placed on a 25.0o incline and friction keeps it from sliding. The coefficient of static friction in this case
vovangra [49]

The frictional force while the mass is sliding will be 46.2 N.

<h3>What is friction force?</h3>

Opposition forces on the surface cause heat loss during the motion of an object known as the friction force.

Given data:

m(mass)= 10.0-kg

Θ (Inclination angle)=25.0o

Coefficient of sliding friction,\rm \mu_k=0.520

Coefficient of static friction,\rm  \mu_s=0.520

The friction force, F=?

Resolve the force in the inclined plane;

\rm F=\mu_s mg cos25^0 \\\\ F=0.520 \times 10 \times 9.81 \times  cos 25 ^0 \\\\ F= 46.2 \ N

Hence, the frictional force while the mass is sliding will be 46.2 N.

To know more about friction force refer to the link;

brainly.com/question/1714663

#SPJ1

5 0
2 years ago
A mass attached to a spring vibrates back and forth. At maximum displacement, the spring force and the
Svetradugi [14.3K]

ANSWER

Velocity of the mass reaches zero

EXPLANATION

We want to identify what hapens to a mass attached toa a spring at maximum displacement.

When a mass attached to a spring is at its maximum position of displacement, the direction of the mass begins to change. This implies that the velocity of the mass will reach zero.

Hence, at maximum displacement, the velocity of the mass reaches zero.

8 0
10 months ago
A 1090 kg car has four 12.7 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to ro
max2010maxim [7]

Answer:

\frac{KE_{Rotational}}{KE_{Total}} = 0.018

Explanation:

To develop this exercise we proceed to use the kinetic energy equations,

In the end we replace

KE_{Total}=KE_{Translational}+KE_{Rotational}

KE_{Total}=\frac{1}{2}m_{car}+4*\frac{1}{2}*I*(\frac{v}{r})^2

Here

I=\frac{1}{2}m_{wheels}*r^2 meaning the 4 wheels,

So replacing

KE_{Rotational}=4\frac{1}{2}*(\frac{1}{2}m_{wheels}*r^2)*(\frac{v}{r})^2=m*v^2

So,

\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}*v^2}{\frac{1}{2}m_{car}*v^2+m_{wheels}*v^2}

\frac{KE_{Rotational}}{KE_{Total}} = \frac{m_{wheels}}{\frac{1}{2}m_{car}+m_{wheels}}

\frac{KE_{Rotational}}{KE_{Total}} =  \frac{10}{545+10}

\frac{KE_{Rotational}}{KE_{Total}} = 0.018

3 0
3 years ago
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