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3 years ago

8

A pitcher throws a baseball at 45 m/s. The baseball has a mass of 400 grams. Disregarding air resistance, the baseball's momentu

m when it hits the catcher's mitt is a0 kg*m/s.

1 answer:

enyata [817]3 years ago

5 0

So it would be 8.90 m/s kg*m/s

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20 POINTS 20 POINTS On a bright sunny day you decide to take a walk. You begin at your home and walk 1000 meters to an ice cream

Kay [80]

Answer:

A) Total Distance = 2000 m and Total displacement = 0 m

B) Average Speed = 44.44 m/min and Average Velocity = 0 m/min

C) Average Speed = 0.7407 m/s and Average velocity = 0 m/s

Explanation:

A) Distance to reach ice cream shop from home = 1000 meters

Therefore distance to get back home would also be 1000 meters.

Total distance traveled = 1000 + 1000 = 2000 metres

Since journey started at home and ended at home, then total displacement = 0 metres.

B) Average speed = Total distance/total time.

Total time = 10 + 15 + 20 = 45 minutes

Since total distance = 2000 m

Then;

Average speed = 2000/45

Average speed = 44.44 m/min

Average velocity = Total displacement/total time

Average velocity = 0/45 = 0 m/min

C) We now want answers in B to be in m/s.

Total time = 45 minutes.

From conversion, 60 seconds make 1 minute. Thus, 45 minutes = 45 × 60 = 2700 seconds

Thus;

Average Speed = 2000/2700

Average Speed = 0.7407 m/s

Average displacement = 0/2700 = 0 m/s

4 0

3 years ago

You repeated a measurement 6 times and recorded the time using a stop watch: 5.8s, 4.6s, 4.8s, 5.1s, 4.3s, 4.6s. What average va

iVinArrow [24]

Answer and Explanation

Arranging the measured values in increasing order;

4.3s, 4.6s, 4.6s, 4.8s, 5.1s, 5.8s

The two outliers are obviously 4.3s and 5.8s; An outlier is a value in a statistical sample which does not fit a pattern that describes most other data point. Outliers make the average value complicated. So, it is usually better for data to be precise with data points spreading out around a small area.

So, the mean is the average of the four remaining data points after removing the outliers.

Mean = (4.6 + 4.6 + 4.8 + 5.1)/4

Mean = 4.775s

So, the value recorded should be 4.775s, 4.78s or 4.8s depending on the number of decimal places allowed.

QED!

8 0

3 years ago

Write about Archimedes principle

Schach [20]

Answer:

Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially, is equal to the weight of the fluid that the body displaces. Archimedes' principle is a law of physics fundamental to fluid mechanics.

8 0

2 years ago

Read 2 more answers

A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o

blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

$\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}}$ (1)

Where:

$T$ - Period of rotation of the satellite, measured in seconds.

$r$ - Distance of the satellite with respect to the center of the planet, measured in meters.

$G$ - Gravitational constant, measured in newton-square meters per square kilogram.

$M$ - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

$r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}$

$r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} }$ (2)

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ and $T = 25800\,s$ , then the distance of the satellite is:

$r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }$

$r \approx 18.897\times 10^{6}\,m$

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

$F = \frac{G\cdot m\cdot M}{r^{2}}$ (3)

Where:

$m$ - Mass of the satellite, measured in kilograms.

$F$ - Force exerted on the satellite by the Earth, measured in newtons.

If we know that $G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}$ , $M = 6.0\times 10^{24}\,kg$ , $m = 6105\,kg$ and $r \approx 18.897\times 10^{6}\,m$ , then the gravitational force is:

$F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}$

$F = 6841.905\,N$

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0

3 years ago

A 2500 kg car traveling to the north is slowed down uniformly from an initial velocity of 20 m/s by a 5620 N braking force actin

Dennis_Churaev [7]

Answer:

the distance traveled by the car is 42.98 m.

Explanation:

Given;

mass of the car, m = 2500 kg

initial velocity of the car, u = 20 m/s

the braking force applied to the car, f = 5620 N

time of motion of the car, t = 2.5 s

The decelaration of the car is calculated as follows;

-F = ma

a = -F/m

a = -5620 / 2500

a = -2.248 m/s²

The distance traveled by the car is calculated as follows;

s = ut + ¹/₂at²

s = (20 x 2.5) + 0.5(-2.248)(2.5²)

s = 50 - 7.025

s = 42.98 m

Therefore, the distance traveled by the car is 42.98 m.

5 0

2 years ago