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slega [8]
3 years ago
5

The batter swings and misses the 40 m/s (90 mph) fastball, and the ball (mass 150 grams) ends up at rest in the catcher's mitt.

How much work does the catcher perform on the ball
Physics
1 answer:
Nadusha1986 [10]3 years ago
5 0

Answer:

The work done by the catcher is 120 J.

Explanation:

Given;

velocity of the fastball, v = 40 m/s

mass of the fastball, m = 150 g = 0.15 kg

Based on work-energy theorem, the work done by the catcher is equal to the kinetic energy of the fastball.

The kinetic energy of the fastball is given as;

K.E = ¹/₂mv²

K.E = ¹/₂ x 0.15 x 40²

K.E = 120 J

Therefore, the work done by the catcher is 120 J.

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A vehicle moving with a uniform a acceleration of 2m/s and has a velocity of 4m/s at a certain time .
Evgesh-ka [11]

Answer:

1 second later the vehicle's velocity will be:

v(1)= 6\,\,\frac{m}{s} \\

5 seconds later the vehicle's velocity will be:

v(5)=14\,\,\frac{m}{s}

Explanation:

Recall the formula for the velocity of an object under constant accelerated motion (with acceleration "a"):

v(t)=v_0+a\,t

Therefore, in this case v_0=4\,\,\frac{m}{s}  and a=2\,\,\frac{m}{s^2}

so we can estimate the velocity of the vehicle at different times just by replacing the requested "t" in the expression:

v(t)=v_0+a\,t\\v(t)=4+2\,\,t\\v(1)=4+2\,(1) = 6\,\,\frac{m}{s} \\v(5)=4+2\,(5)=14\,\,\frac{m}{s}

3 0
3 years ago
If a body p with a positive charge is placed in contact with a body q (initially uncharged), what will be the nature of the char
uysha [10]

If a body p with a positive charge is placed in contact with a body q (initially uncharged), then the nature of charge gained by q must be positive, because rubbing an uncharged body with a charged body or placed in contact with a positive charged body, helps gain a charge to the uncharged body.

There are a variety of methods to charge an object. One method is known as induction. In the induction process, a charged object is brought near but not touched to a neutral conducting object.

Let's know, how a element gain positive charge?

A positive charge occurs when the number of protons exceeds the number of electrons. A positive charge may be created by adding protons to an atom or object with a neutral charge. A positive charge also can be created by removing electrons from a neutrally charged object.

To learn more about Positive charge here

brainly.com/question/2903220

#SPJ4

5 0
2 years ago
Relationship between frequency, amplitude, wavelength and energy in a transverse wave.
liq [111]
In a transverse wave:
- Oscillations are perpendicular to the direction of energy travelling
- Frequency is the amount of complete waves passing a certain point in one second (measured in hertz, Hz)
- Wavelength is the distance from any point on one wave to the same point on the following wave
- The amplitude is the maximum displacement of the particles from their average position (and be measured from the horizontal mid-point of the wave to either the peak or trough)

There isn't always a defined relationship between these features. However, frequency × wavelength = velocity of the wave.
3 0
4 years ago
13) A mass attached to the free end of a spring executes simple harmonic motion according to the equation y = (0.50 m) sin (18π
Kitty [74]

Hi there!

The period is given by:

T = \frac{2\pi}{w}

T = Period (sec)

w = angular frequency (rad/sec)

According to the equation for SHM in terms of position:

y(t) = Asin(ωt + φ)

A = Amplitude (m)

ω = angular frequency (rad/sec)

t = time (sec)

φ = phase angle

In this instance, the angular frequency is given as 18π.

Plug this value into the equation for T:

T= \frac{2\pi}{18\pi} = \frac{1}{9} = \boxed{0.111 s}

4 0
3 years ago
A box has a weight of 150 N and is being pulled across a horizontal floor by a force that has a magnitude of 110 N. The pulling
ivann1987 [24]

Answer:

42.99°

Explanation:

F_h = Kinetic friction force

F_{\theta} = Pulling force at angle \theta

N_h = Weight of the box = 150 N

Kinetic friction force

F_h=\muN_h

Pulling force at angle \theta

F_{\theta}=\muN_{\theta}

N = Pulling force

According to question

\frac{F_h}{F_{\theta}}=\frac{2}{1}\\\Rightarrow \frac{\muN_h}{\muN_{\theta}}=2\\\Rightarrow \frac{N_h}{N_{\theta}}=2\\\Rightarrow N_{\theta}=\frac{N_h}{2}\\\Rightarrow N_{\theta}=\frac{150}{2}\\\Rightarrow N_{\theta}=75\ N

Applying Newton's second law in the vertical direction we get

N_h-Nsin\theta=N_{\theta}\\\Rightarrow 150-110sin\theta=75\\\Rightarrow \theta=sin^{-1}\frac{75}{110}\\\Rightarrow \theta=42.99\ ^{\circ}

The angle is 42.99°

8 0
4 years ago
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