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Digiron [165]
3 years ago
14

Which object has the least momentum? object a: m = 1 kg, v = 100 m/s object b: m = 10 kg, v = 12 m/s object c: m = 0.5 kg, v = 1

000 m/s object d: m = 100 kg, v = 2 m/s?
Physics
1 answer:
nirvana33 [79]3 years ago
8 0
A = 1*100 = 100 Ns
b = 10 * 12 = 120 Ns
c = 0.5*1000 = 500 Ns
d = 100 * 2 = 200 Ns

a has least momentum
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Not sure if it went through last time. Please help asap!
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3 years ago
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A 50 kg skydiver is falling downwards and accelerating 6 m/s2 down. What is the net force on the skydiver?
Montano1993 [528]

Net Force = (mass) x (acceleration)  (Newton #2)

Net Force = (50 kg) x (6 m/s² down)

Net Force = (50 * 6) (kg-m/s² down)

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6 0
2 years ago
A sled of mass 2.12 kg has an initial speed of 5.49 m/s across a horizontal surface. The coefficient of kinetic friction between
Darya [45]

Answer:

The speed of the sled is 3.56 m/s

Explanation:

Given that,

Mass = 2.12 kg

Initial speed = 5.49 m/s

Coefficient of kinetic friction = 0.229

Distance = 3.89 m

We need to calculate the acceleration of sled

Using formula of acceleration

a = \dfrac{F}{m}

Where, F = frictional force

m = mass

Put the value into the formula

a=\dfrac{\mu mg}{m}

a=\mu g

a=0.229\times9.8

a=2.244\ m/s^2

We need to calculate the speed of the sled

Using equation of motion

v^2=u^2-2as

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value in the equation

v ^2=(5.49)^2-2\times2.244\times3.89

v=\sqrt{(5.49)^2-2\times2.244\times3.89}

v=3.56\ m/s

Hence, The speed of the sled is 3.56 m/s.

8 0
3 years ago
Five identical quintuplets leave earth when they reach the age of 21, in the year 2121. Each quintuplet goes on a spaceship jour
Elena-2011 [213]

Answer:

Explanation:

This is a problem based on time dilation , a theory given by Albert Einstein .

The formula of time dilation is as follows .

t₁ = \frac{t}{\sqrt{1-\frac{v^2}{c^2} } }

t is time measured on the earth and t₁ is time measured by man on ship .

A ) Given t = 20 years , t₁ = ? v = .4c

\frac{20}{\sqrt{1-\frac{.16c^2}{c^2} } }

=1.09 x 20

t₁= 21.82 years

B ) Given t = 5 years , t₁ = ? v = .2c

\frac{5}{\sqrt{1-\frac{.04c^2}{c^2} } }

=1.02 x 5

t₁= 5.1 years

C ) Given t = 10 years , t₁ = ? v = .8c

\frac{10}{\sqrt{1-\frac{.64c^2}{c^2} } }

=1.67 x 10

t₁= 16.7  years

D ) Given t = 10 years , t₁ = ? v = .4c

\frac{10}{\sqrt{1-\frac{.16c^2}{c^2} } }

=1.09 x 10

t₁= 10.9  years

E ) Given t = 20 years , t₁ = ? v = .8c

\frac{20}{\sqrt{1-\frac{.64c^2}{c^2} } }

=1.67 x 20

t₁= 33.4   years

7 0
2 years ago
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