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3 years ago

0.0010 kg pellet is fired at a speed of 52.2 m/s at a motionless 3.3 kg piece of balsa wood. When the pellet

Physics

1 answer:

Paraphin [41]3 years ago

7 0

The formula for momentum is p = m*v

The conservation of momentum suggests:

m*vi = m*vf (initial mass times initial velocity = final mass times final velocity or initial momentum = final momentum)

(0.0010)(52.2) = (0.0010 + 3.3)vf

vf = (0.0010)(52.2)/(0.0010 + 3.3) = 0.0522/3.301 ≈ 0.01581 m/s

To the nearest thousandth ≈ .016 m/s

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two cars are moving at constant speeds in a straight line along a major highway. The first is travelling at 20ms^-1 and the seco

Reika [66]

Answer:

$t=750s$

Explanation:

The two cars are under an uniform linear motion. So, the distance traveled by them is given by:

$\Delta x=vt\\x_f-x_0=vt\\x_f=x_0+vt$

$x_f$ is the same for both cars when the second one catches up with the first. If we take as reference point the initial position of the second car, we have:

$x_0_1=6km\\x_0_2=0$

We have $x_f_1=x_f_2$ . Thus, solving for t:

$x_0_1+v_1t=x_0_2+v_2t\\x_0_1=t(v_2-v_1)\\t=\frac{x_0_1}{v_2-v_1}\\t=\frac{6*10^3m}{28\frac{m}{s}-20\frac{m}{s}}\\t=750s$

8 0

2 years ago

Some birds migrate 20,000 miles. If 1 mile equals 1.6 kilometers, calculate the distance these birds fly in kilometers.

faust18 [17]

Set up a proportion

1 mile/1.6 km = 20,000miles/x

Cross Multiply
x = (20,000) * (1.6)
x = 32,000 kilometers

6 0

3 years ago

Read 2 more answers

A particle of mass 4.00 kg is attached to a spring with a force constant of 100 N/m. It is oscillating on a frictionless, horizo

zloy xaker [14]

Solution :

Given :

Mass attached to the spring = 4 kg

Mass dropped = 6 kg

Force constant = 100 N/m

Initial amplitude = 2 m

Therefore,

a). $v_{initial} = A w$

$= 2 \times \sqrt{\frac{100}{4}}$

= 10 m/s

Final velocity, v at equilibrium position, v = 5 m/s

Now, $\frac{1}{2}(4+4)5^2 = \frac{1}{2} kA'$

A' = amplitude = 1.4142 m

b). $T=2 \pi \sqrt{\frac{m}{k}}$

m' = 2m

Hence, $T'=\sqrt2 T$

c). $\frac{\frac{1}{2}(4+4)5^2 + \frac{1}{2}\times 4 \times 10^2}{\frac{1}{2} \times 4 \times 10^2}$

$=\frac{1}{2}$

Therefore, factor $=\frac{1}{2}$

Thus, the energy will change half times as the result of the collision.

7 0

2 years ago

A positive charge is moved from point A to point B along an equipotential surface. How much work is performed or required in mov

Nitella [24]

Answer:

No work is performed or required in moving the positive charge from point A to point B.

Explanation:

Lets take

Q= Positive charge which move from point A to point B along

Voltage difference,ΔV =V₁ - V₂

The work done

W = Q . ΔV

Given that charge is moved from point A to point B along an equipotential surface.It means that voltage difference is zero.

ΔV = 0

W = Q . ΔV

W = Q x 0

W= 0 J

So work is zero.

5 0

2 years ago

The apparent weight of a student in alift is 564N . if the mass of the student is 60.3kg, what is the acceleration of the lift ?

Yuki888 [10]

Answer:

-.457 m/s^2

Explanation:

Actual weight = 60 .3 (9.81) = 591.54 N

Accel of lift changes this to 60.3 ( 9.81 - L) where L - accel of lift

60.3 ( 9.81 - L ) = 564

solve for L = .457 m/s^2 DOWNWARD

so L = - .457 m/s^2

4 0

1 year ago