If a 100 N block is resting on a steel table with a coefficient of

A ball is thrown vertically down from the edge of a cliff with a speed of 4 m/s,

Serhud [2]

By using the equation speed = distance/time we can solve for distance. The speed is 4 m/s and the time is 12 seconds. We need to rearrange the equation to Speed * Time = distance. 4(12) = 48; 48 = distance. The cliff is 48 meters high.

3 0

3 years ago

A 4.30 g bullet moving at 943 m/s strikes a 730 g wooden block at rest on a frictionless surface. The bullet emerges, traveling

Ymorist [56]

Answer:

(a)2.7 m/s

(b) 5.52 m/s

Explanation:

The total of the system would be conserved as no external force is acting on it.

Initial momentum = final momentum

⇒(4.30 g × 943 m/s) + (730 g × 0) = (4.30 g × 484 m/s) + (730 g × v)

⇒ 730 ×v = (4054.9 - 2081.2) =1973.7

⇒v=2.7 m/s

Thus, the resulting speed of the block is 2.7 m/s.

(b) since, the momentum is conserved, the speed of the bullet-block center of mass would be constant.

$V_{COM} = \frac{m_b}{m_b+m_{bl}}v_{bi}=\frac{4.30}{4.30+730}\times 943 m/s = 5.52 m/s$

Thus, the speed of the bullet-block center of mass is 5.52 m/s.

4 0

4 years ago

A light bulb has a power rating of 60 W. What is the current running into it if it is plugged into a 120 V outlet?

kifflom [539]

Answer:

0.5 Amperes

Explanation:

Information that we have:

Power: $P=60W$

Voltage: $V=120V$

and we use the equation that relates this two quantities to the Current (I):

$I=\frac{P}{V}$

where I is the curent, P is the Power and V is the Voltage.

we substitute the values given for the Power and the Voltage to find the current:

$I=\frac{60W}{120V}\\ I=0.5A$

the Current running into the bulb is 0.5 Amperes

3 0

3 years ago

A railroad car of mass 2.00 3 104 kg moving at 3.00 m/s collides and couples with two coupled railroad cars, each of the same ma

FrozenT [24]

Given:

Mass of the rail road car, m = 2 kg

velocity of the three cars coupled system, v' = 1.20 m/s

velocity of first car, $v_{a}$ = 3 m/s

Solution:

a) Momentum of a body of mass 'm' and velocity 'v' is given by:

p = mv

Now for the coupled system according to law of conservation of momentum, total momentum of a system before and after collision remain conserved:

$mv_{a} + 2mv_{b} = (m + 2m)v'$ (1)

where,

$v_{a}$ = velocity of the first car

$v_{b}$ = velocity of the 2 coupled cars after collision

Now, from eqn (1)

$v' = \frac{v_{a} + 2v_{b}}{3}$

$v' = \frac{3.00 + 2\times 1.20}}{3}$

v' = 1.80 m/s

Therefore, the velocity of the combined car system after collision is 1.80 m/s

7 0

3 years ago

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An ideal parallel plate capacitor with a cross-sectional area of 0.4 cm2 contains a dielectric with a dielectric constant of 4 a

Mrrafil [7]

Answer: $283.2\times 10^{-9}\ nC$

Explanation:

Given

Cross-sectional area $A=0.4\ cm^2$

Dielectric constant $k=4$

Dielectric strength $E=2\times 10^8\ V/m$

Distance between capacitors $d=5\ mm$

Maximum charge that can be stored before dielectric breakdown is given by

$\Rightarrow Q=CV\\\\\Rightarrow Q=\dfrac{k\epsilon_oA}{d}\cdot (Ed)\quad\quad [V=E\cdot d]\\\\\Rightarrow Q=k\epsilon_oAE\\\\\Rightarrow Q=4\times 8.85\times 10^{-12}\times 0.4\times 10^{-4}\times 2\times 10^8\\\\\Rightarrow Q=28.32\times 10^{-8}\\\\\Rightarrow Q=283.2\times 10^{-9}\ nC$

4 0

3 years ago

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