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zhenek [66]
3 years ago
7

Suppose you are driving a car and your friend, who is with you in the car, tosses a softball up and down from her point of view.

A second friend of yours stands on the street and sees you passing by and says, "You are testing projectile motion with the softball!". Your friend on the street claims that each of you measures a different value for the vertical component of the initial velocity of the softball. Do you agree with the statement?
Physics
1 answer:
Sholpan [36]3 years ago
3 0

Answer:

No, i disagree.

Explanation:

If the car is moving, it only has a velocity with a component in the horizontal direction. If we use galilean relativity, the velocity of the ball observed by my friend standing in the ground should only be affected in the horizonal direction, while the vertical stays the same for both observers.

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How is newtons law of motion used for testing the safety of automobile
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Newton's first law of motion is an object in motion stays in motion until acted upon by another force. Driving at 30 mph in a car is going to stay constant until you crash the car into a wall, stopping the car.
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If you weigh 38 kilograms on your bathroom scale, your weight in space will be ________.
harina [27]

Answer:

Less than 36kilo's

Explanation:

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3 years ago
The equation below is for potassium oxide.
Reika [66]
The answer is C 4:1.
7 0
3 years ago
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43 kg bear slides, from rest, 15 m down a lodgepole pine tree, moving with a speed of 5.5 m/s just before hitting the ground. (a
olga2289 [7]

Answer:

-6327.45 Joules

650.375 Joules

378.47166 N

Explanation:

h = Height the bear slides from = 15 m

m = Mass of bear = 43 kg

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity of bear = 5.5 m/s

f = Frictional force

Potential energy is given by

P=mgh\\\Rightarrow P=43\times -9.81\times 15\\\Rightarrow P=-6327.45\ J

Change that occurs in the gravitational potential energy of the bear-Earth system during the slide is -6327.45 Joules

Kinetic energy is given by

K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 43\times 5.5^2\\\Rightarrow K=650.375\ J

Kinetic energy of the bear just before hitting the ground is 650.375 Joules

Change in total energy is given by

\Delta E=fh=-(\Delta K+\Delta P)\\\Rightarrow fh=-(650.375-6327.45)\\\Rightarrow fh=5677.075\\\Rightarrow f=\frac{5677.075}{h}\\\Rightarrow f=\frac{5677.075}{15}\\\Rightarrow f=378.47166\ N

The frictional force that acts on the sliding bear is 378.47166 N

5 0
3 years ago
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A proton is released from rest at the positive plate of a parallel-plate capacitor. It crosses the capacitor and reaches the neg
katrin2010 [14]

Answer:

2.1x10^6m/s

Explanation:

One electron has a charge of –1.602e-19 C

mass of electron is 9.1e-31 kg

mass of proton is 1.6726e−27 kg

mass ratio is 1.6726e−27 / 9.1e-31 = 1838

The force is constant, F

distance is constant, d

a = F/m

a increases by a factor 1838, as m decreases by that factor

a = a₀1838

v₀² = 2a₀d

v² = 2a₀d1838

v²/v₀² = 2a₀d1838 / 2a₀d = 1838

v² = 1838v₀² = 1838(45000)²

v = 45000√1838 = 2.1e6 m/s

5 0
3 years ago
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