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belka [17]
3 years ago
9

A 12.0-g bullet is fired horizontally into a 112-g wooden block that is initially at rest on a frictionless horizontal surface a

nd connected to a spring having spring constant 149 N/m. The bullet becomes embedded in the block. If the bullet-block system compresses the spring by a maximum of 83.5 cm, what was the speed of the bullet at impact with the block
Physics
1 answer:
sergiy2304 [10]3 years ago
7 0
The speed would be in a decimal? Or do you want it in a fraction?
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Science Mixtures Question( giving brainly, thanks, and rank of 5 stars.)
vazorg [7]

Answer:

Condensation

Explanation:

That is when water from the air collects as a drop. also can be rain.

4 0
3 years ago
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The brakes of an automobile can decelerate it at -4.7m/s^2. If the automobile is traveling at the rate of 31.0m/s How far will i
OleMash [197]

Answer:

204.5 m

Explanation:

a = v \ t

t = v \ a

t = 31 \ 4.7 = 6.6s

v = d \ t

d = v * t

d = 31 * 6.6

d= 204.5 meters

6 0
3 years ago
A motorcycle is following a car that is traveling at constant speed on a straight highway. Initially, the car and the motorcycle
Sunny_sXe [5.5K]

Answer:

a) 5.09 seconds

b) 107.07 meters

Explanation:

a) As we know

t_2- t_1 = \sqrt{\frac{2 X}{a} }

Substituting the given values we get

t_2 - t_1 = \sqrt{\frac{2 * 52}{4} } \\t_2 - t_1 = 5.09

It takes 5 .09 s for the motorcycle to accelerate until it catches up with the car

b)

X_{t`2} = v_i \sqrt{\frac{2X}{a} } + 0.5 a\sqrt{\frac{2X}{a} }\\X_{t`2} =  (v_i + 0.5 a) \sqrt{\frac{2X}{a} }\\X_{t`2} =  ( 19 + 2)  \sqrt{\frac{2* 52}{4} }\\X_{t`2} =  21 * 5.09\\X_{t`2} = 107.07

4 0
3 years ago
A rectangular wire loop is pulled out of a region of uniform magnetic field B at a constant speed v. What is true about the indu
Paul [167]

Answer:

There is a constant emf induced in the loop.

Explanation:

In the uniform magnetic field suppose the rectangular wire loop of length L and width b is moved out with a uniform velocity v. suppose any instance x length of the loop is out of the magnetic field and L-x length is inside the loop.

Area of loop outside the field = b(L-x)

we know that flux φ= BA

B= magnitude of magnetic field , A=  area

and emf \epsilon= \frac{d\phi}{dt}

\epsilon=B\frac{dA}{dt}

\epsilon=B\frac{db(L-x)}{dt}

\epsilon=Bb\frac{d(L-x)}{dt}

B,b and L are constant and dx/dt = v

⇒ε = -Bbv

which is a constant hence There is a constant emf induced in the loop.

5 0
4 years ago
If Superman wants to slow down a fast car with speed 20 m/s and mass 1000 kg, how much force in N does he need to apply if he wa
saveliy_v [14]

Answer:

Force of 6250N is required to slow down the body to zero.

Explanation:

Speed(V)=20m/s

Initial speed (u) =0m/s

Mass(M)=1000kg

Time(t)=3.2seconds

Using first equation of motion to find the acceleration of the body:

V=U + at

20m/s=0 + a*3.2s

Therefore, a=(20m/s)/3.2s

a=6.25m/s^2

To find the force required to stop the body or get it down to zero:

Force(F)=mass x acceleration

F=(1000kg)*(6.25m/s^2)

F=6250N.

Therefore a force of 6250N is required to slow it down.

8 0
3 years ago
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