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iren2701 [21]
3 years ago
8

SOMEONE HELP ME PLEASE

Physics
2 answers:
TiliK225 [7]3 years ago
6 0
Newton’s second law of motion is F=ma
mihalych1998 [28]3 years ago
5 0

Explanation:

By the second law of Newton we get the relation

F = ma

You might be interested in
A bullet is shot vertically upward with an initial velocity of 128 ft/s. The bullet's height after t seconds is y(t) = 128t - 16
saul85 [17]

The height of the bullet when the velocity is zero is 256 ft.

<h3>Height of the bullet when the velocity is zero </h3>

The height of the bullet when the velocity is zero is determined by taking derivative of the function as shown below;

v = \frac{dy}{dt} = 128 -32t \\\\when \ v \ is \ zero\\\\v = 0\\\\128 - 32t = 0\\\\32t = 128\\\\t = \frac{128}{32} \\\\t = 4 \  s

The height of the bullet at this time is calculated as follows;

y(4) = 128(4) - 16(4)^2\\\\y(4) = 256 \ ft

Learn more about height of projectiles here: brainly.com/question/10008919

6 0
3 years ago
Which year was pluto no longer considered a planet?.
katrin2010 [14]
2006, i hope this helps
6 0
2 years ago
A 75kg hockey player is skating across the ice at a speed of 6.0m/s. What is the magnitude of the average force required to stop
liq [111]

Answer:

692.31 N

Explanation:

Applying,

F = ma............... Equation 1

Where F = Average force required to stop the player, m = mass of the player, a = acceleration of the player

But,

a = (v-u)/t............ Equation 2

Where v = final velocity, u = initial velocity, t = time.

Substitute equation 2 into equation 1

F = m(v-u)/t............ Equation 3

From the question,

Given: m = 75 kg, u = 6.0 m/s, v = 0 m/s (to stop), t = 0.65 s

Substitute these values into equation 3

F = 75(0-6)/0.65

F = -692.31 N

Hence the average force required to stop the player is 692.31 N

6 0
3 years ago
Suppose the coefficient of static friction between the road and the tires on a car is 0.638 and the car has no negative lift. Wh
larisa [96]

Answer:

12.6332454263 m/s

Explanation:

m = Mass of car

v = Velocity of the car

\mu = Coefficient of static friction = 0.638

g = Acceleration due to gravity = 9.81 m/s²

r = Radius of turn = 25.5 m

When the car is on the verge of sliding we have the force equation

\dfrac{mv^2}{r}=\mu mg\\\Rightarrow v=\sqrt{\mu gr}\\\Rightarrow v=\sqrt{0.638\times 9.81\times 25.5}\\\Rightarrow v=12.6332454263\ m/s

The speed of the car that will put it on the verge of sliding is 12.6332454263 m/s

4 0
3 years ago
I’ll give u BRAINLIEST PLEASE!! HURRY
zalisa [80]

Answer:

C

Explanation:

6 0
3 years ago
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