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qaws [65]
3 years ago
15

A container can hold a mass of 40 grams. If Jackson has a fake gold chain that has a density of 10 g/cm3 that occupies a space o

f 5 cm3, will the box be able to hold it? (D=M/V, V=M/D, M=DxV)
Physics
1 answer:
eimsori [14]3 years ago
5 0
Use the formula M=D×V:

M=10 g/cm³ * 5 cm³ = 50 g

which is more than 40 grams, so the container cannot hold the chain.
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A 42.0-kg parachutist is moving straight downward with a speed of 3.85 m/s. (a) If the parachutist comes to rest with constant a
RideAnS [48]

Answer:

-414.96 N

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{0^2-3.85^2}{2\times 0.75}\\\Rightarrow a=-9.88\ m/s^2

F=ma\\\Rightarrow F=42\times -9.88\\\Rightarrow F=-414.96\ N

The force the ground exerts on the parachutist is -414.96 N

If the distance is shorter than 0.75 m then the acceleration will increase causing the force to increase

5 0
3 years ago
An asteroid with a mass of 5.0 x 105 kg collides with the Earth and slides horizontally along the ground without bouncing (not a
V125BC [204]
The strength of the friction doesn't matter. Neither does the distance or the time the asteroid takes to stop. All that matters is that the asteroid has

1/2 (mass) (speed squared)

of kinetic energy when it lands, and zero when it stops.
So

1/2 (mass) (original speed squared)

is the energy it loses to friction in order to come to rest.
8 0
3 years ago
Which is correct? I need to know!!
Mnenie [13.5K]
What website are you using?
6 0
3 years ago
Two ping pong balls have been painted with metallic paint and charged by contact with an Van de Graaff generator. The charge on
andre [41]

Answer:

0.035 N

Explanation:

Parameters given:

Charge q1 = -3.31x10^(-7) C

Charge q2 = -5.7x10^(-7) C.

Distance between them, R = 22 cm = 0.22 m

Electrostatic force between to particles is given as:

F = (k* q1 * q2) / R²

F = (9 * 10^9 * -3.31 * 10^(-7) * -5.7 * 10^(-7)) / 0.22²

F = 0.035 N

6 0
3 years ago
Read 2 more answers
A child is riding a merry-go-round that has an instantaneous angular speed of 12 rpm. If a constant friction torque of 12.5 Nm i
sammy [17]

Answer:

-0.25 rad/s^2

Explanation:

The equivalent of Newton's second law for rotational motions is:

\tau = I \alpha

where

\tau is the net torque applied to the object

I is the moment of inertia

\alpha is the angular acceleration

In this problem we have:

\tau = -12.5 Nm (net torque, with a negative sign since it is a friction torque, so it acts in the opposite direction as the motion)

I=50.0 kg m^2 is the moment of inertia

Solving for \alpha, we find the angular acceleration:

\alpha = \frac{\tau}{I}=\frac{-12.5 Nm}{50.0 kg m^2}=-0.25 rad/s^2

3 0
3 years ago
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