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3 years ago

Pls help me idk what to do

Physics

1 answer:

Alika [10]3 years ago

6 0

The dot would be the earthquake epicenter. That is where the earthquake originated.

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A 10 g bullet is fired into, and embeds itself in, a 2 kg block attached to a spring with a force constant of 19.6 n/m and whose

dmitriy555 [2]

Answer:

$x=0.478\ m$ is the compression in the spring

Explanation:

Given:

mass of the bullet, $m=10\ g=0.01\ kg$

mass of block, $M=2\ kg$

stiffness constant of the spring, $k=19.6\ N.m^{-1}$

initial velocity of the spring just before it hits the block, $u=300\ m.s^{-1}$

<u>Now since the bullet-mass gets embed into the block, we apply the conservation of momentum as:</u>

$m.u=(M+m).v$

$0.01\times 300=(2+0.01)\times v$

$v=1.4925\ m.s^{-1}$

Now this kinetic energy of the combined mass gets converted into potential energy of the spring.

$\rm Kinetic\ energy=Spring\ potential\ energy$

$\frac{1}{2} (M+m).v^2=\frac{1}{2} k.x^2$

$\frac{1}{2} \times (2+0.01)\times 1.4925^2=\frac{1}{2} \times 19.6\times x^2$

$x=0.478\ m$ is the compression in the spring

4 0

3 years ago

Read 2 more answers

A trapeze artist, with swing, weighs 800 N; he is momentarily heldto one side by his partner using a horizontal force so that th

Tamiku [17]

Answer:

461.88 N

Explanation:

$F_{g}$ = Weight of the swing = 800 N

$T$ = Tension force in the rope

$F$ = Horizontal force being applied by the partner

Using equilibrium of force in vertical direction using the force diagram, we get

$T Cos30 = F_{g}\\T Cos30 = 800\\T = \frac{800}{Cos30} \\\\T = 923.76 N$

Using equilibrium of force in horizontal direction using the force diagram, we get

$F = T Sin30\\F = (923.76) (0.5)\\F = 461.88 N$

6 0

2 years ago

A car of mass 500kg travelling at 60m/s has it speed reduced to 40m/s by a constant breaking force over a distance of 200m. find

uranmaximum [27]

Answer:

Ek1 = 900000 [J]

Ek1 = 400000 [J]

Explanation:

In order to solve this problem we must remember that kinetic energy is defined as the product of mass by velocity squared by a medium. Therefore using the following equation we have:

$E_{k1}=\frac{1}{2}*m*v1^{2}$

where:

m = mass = 500 [kg]

v1 = 60 [m/s]

So we have:

Ek1 = 0.5*500*(60^2)

Ek1 = 900000 [J]

and:

Ek2 = 0.5*500*(40^2)

Ek2 = 400000 [J]

6 0

3 years ago

An upward moving object must be experiencing (or at least usually does experience) an upward force.

gayaneshka [121]

Gravity pulls objects down to the earth

8 0

2 years ago

An electric field from a charge has a magnitude of 1.5 × 104 N/C at a certain location that points inward. If another charge wit

zysi [14]

Answer:

-0.045 N, they will attract each other

Explanation:

The strength of the electrostatic force exerted on a charge is given by

$F=qE$

where

q is the magnitude of the charge

E is the electric field magnitude

In this problem,

$q=3.0\cdot 10^{-6}C$

$E=-1.5\cdot 10^4 N/C$ (negative because inward)

So the strength of the electrostatic force is

$F=(-3.0\cdot 10^{-6}C)(1.5\cdot 10^4 N/C)=-0.045 N$

Moreover, the charge will be attracted towards the source of the electric field. In fact, the text says that the electric field points inward: this means that the source charge is negative, so the other charge (which is positive) is attracted towards it.

6 0

3 years ago

Read 2 more answers