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2 years ago

Date

Physics

1 answer:

shtirl [24]2 years ago

4 0

Answer:

W=F×H,Power=W/Time

Explanation:

W=Force ×Height

750N×16 =12000

Power=Work/Time

12000/5=2400J

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4. A driver travels 135 km, east in 1.5 h, stops for 45 minutes for lunch,

stira [4]

Answer:

The driver's average velocity is 82.35 km/h.

Explanation:

Given:

The motion of the driver can be divided into 3 parts:

i. Displacement of the driver in 1.5 hours = 135 km

ii. Rest for 45 minutes.

iii. Displacement in next 2 hours = 215 km

The direction of motion remains same (east).

Now, total displacement of the driver is, $D_{Total}=135+215=350$ km.

Rest time is 45 minutes. Converting it to hours, we need to use the conversion factor $1\textrm{ min} = \frac{1}{60}$ hour.

So, 45 minutes in hours is equal to $\frac{45}{60}=0.75$ hours.

Now, total time taken for the complete journey is, $\Delta t=1.5+\frac{45}{60}+2=1.5+0.75+2=4.25\textrm{ h}$

Average velocity is given as:

$v_{avg}=\frac{\textrm{Total displacement}}{Total time}=\frac{350}{4.25}=82.35\textrm{ km/h}$

Therefore, the driver's average velocity is 82.35 km/h

4 0

2 years ago

When an object is lifted against the force of gravity it has gravitational potential energy. The energy depends on the object's

USPshnik [31]

The answer is actually True, I just took the test and it was correct.

5 0

3 years ago

Read 2 more answers

How to draw the ray diagram? Pls answer

weqwewe [10]

1. Pick a point on the top of the object and draw two incident rays traveling towards the mirror. Using a straight edge, accurately draw one ray so that it passes exactly through the focal point on the way to the mirror. Draw the second ray such that it travels exactly parallel to the principal axis.

7 0

3 years ago

A capacitor is charged until it holds 5.0 j of energy. it is then connected across a 10-kω resistor. in 13.6 ms , the resistor d

prohojiy [21]

Answer:

The capacite is C=5.32 uF using the equations of voltage and energy in capacitance

Explanation:

The energy holds is 5 J and the resistor dissipates 2J so the energy total is 3J

Using:

$V_{t}= V_{o}e^{\frac{-t}{R*C} }$

Voltage in this case is the energy dissipated so

$E_{t}= E_{o}e^{\frac{-t}{R*C} }$

$\frac{\sqrt{E_t} }{\sqrt{E_o} } = e^{\frac{-t}{R*C} }$

$\frac{\sqrt{3 J} }{\sqrt{5J} } = e^{\frac{-13.6ms}{10kw*C} }$

Using the equation to find capacitance

$ln 0.775= e^{\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\ln(0.775)= ln * e^{\frac{-13.6 x10^{3s} }{10x10^{3}*C }} \\\\ln(0.775)= {\frac{-13.6 x10^{3} }{10x10^{3}*C }} \\C= \frac{-13.6 x10^{-3} }{10x10x^{3}*ln(0.775) }$