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2 years ago

What is the momentum of a 100-kilogram fullback carrying a football on a play at a velocity of 3.5. m/sec

Physics

2 answers:

podryga [215]2 years ago

7 0

Answer: 350 kg m/s

Explanation: Momentum can be defined as "mass in motion."

P=MV

P = momentum

M=mass

V=velocity

In the question, we were given;

M= 100-kilogram

V= 3.5. m/sec

Therefore,

Momentum= Mass * velocity

Momentum= 100kg * 3.5 m/sec

Momentum= 350kg.m/sec.

Olenka [21]2 years ago

6 0

Answer:

350 kg m/s

Explanation:

Momentum = mass × velocity

p = mv

p = (100 kg) (3.5 m/s)

p = 350 kg m/s

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Large amplitude of sound vibrations will produce.....

user100 [1]

Answer:

louder sound.

Explanation:

The amplitude of a sound wave determines its loudness or volume. A larger amplitude means a louder sound

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3 years ago

PLZ HELP WILL GIVE BRAINLIEST

xxTIMURxx [149]

Answer:

12.7m/s

Explanation:

Given parameters:

Mass of the diver = 77kg

Height = 8.18m

Unknown:

Final velocity = ?

Solution:

To solve this problem, we use one of the motion equations.

v² = u² + 2gh

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

h is the height

v² = 0² + (2 x 9.8 x 8.18)

v² = 160.3

v = 12.7m/s

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2 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago

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Klio2033 [76]

Answer:

28.15

Explanation:

the ateps is in the photo

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3 years ago

Need help y’all ASAP please...physics

dolphi86 [110]

Answer:

t = 3/8 seconds

Explanation:

h=-16t^2 - 10t+6

h= 0 when it hits the ground

0=-16t^2 - 10t+6

factor out a -2

0= -2(8t^2 +5t -3)

divide by -2

0 = (8t^2 +5t -3)

factor

0=(8t-3) (t+1)

using the zero product property

8t-3 = 0 t+1 =0

8t = 3 t= -1

t = 3/8 t= -1

t cannot be negative ( no negative time)

t = 3/8 seconds

3 0

2 years ago