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3 years ago

Which of the following is true about the following lever?

Physics

2 answers:

NISA [10]3 years ago

7 0

Answer:

Explanation:

i think the answer is c... don't quote me on that... i'm not 100% sure...

Natali [406]3 years ago

4 0

Oh jk haha LOL bye hahaha

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You have 1.2kg of (AU)gold (that's quite a bit of money at $1839/ounce). Using a heat source you apply 3096 J to the gold and re

melamori03 [73]

Answer:

129 J/Kg°C

Explanation:

Given :

Mass of gold, m = 1.2kg

Quantity of heat applied, Q = 3096 J

Temperature, t2 = 40°C

Temperature, t1 = 20°C

Change in temperature, dt = (40-20)°C = 20°C

Using the relation :

Q = mcdt

Where, C = specific heat capacity of gold

3096 = 1.2kg * C * 20°C

3096 J = 24kg°C * C

C = 3096 J / 24 kg°C

C = 129 J/Kg°C

7 0

2 years ago

Compute the kinetic energy of a proton (mass 1.67×10−27kg ) using both the nonrelativistic and relativistic expressions for spee

JulsSmile [24]

Answer:

The non-relativistic kinetic energy of a proton is $6.76\times10^{-12}\ J$

The relativistic kinetic energy of a proton is $7.25\times10^{-12}\ m/s$

Explanation:

Given that,

Mass of proton $m=1.67\times10^{-27}\ kg$

Speed $v= 9.00\times10^{7}\ m/s$

We need to calculate the kinetic energy for non relativistic

Using formula of kinetic energy

$K.E=\dfrac{1}{2}mv^2$

Put the value into the formula

$K.E=\dfrac{1}{2}\times1.67\times10^{-27}\times(9.00\times10^{7})^2$

$K.E=6.76\times10^{-12}\ J$

We need to calculate the kinetic energy for relativistic

Using formula of kinetic energy

$K.E=mc^2(\sqrt{(\dfrac{1}{1-\dfrac{v^2}{c^2}})}-1)$

$K.E=1.67\times10^{-27}\times(3\times10^{8})^{2}\cdot\left(\sqrt{\frac{1}{1-\frac{\left(9.00\times10^{7}\right)^{2}}{(3\times10^{8})^{2}}}}-1\right)$

$K.E=7.25\times10^{-12}\ m/s$

Hence, The non-relativistic kinetic energy of a proton is $6.76\times10^{-12}\ J$

The relativistic kinetic energy of a proton is $7.25\times10^{-12}\ m/s$

7 0

2 years ago

A 48.0-kg skater is standing at rest in front of a wall. By pushing against the wall she propels herself backward with a velocit

Kipish [7]

Answer:

F = 47.6 N

Explanation:

Newton's 2nd law can be expressed as the rate of change of the total momentum, respect of time, as follows:

$F = \frac{\Delta p}{\Delta t}$

So, in order to find the average force exerted by the skater on the wall, we can find the change in momentum due to the force exerted by the wall (which is equal and opposite to the one exerted by the skater), and divide it by the time interval , as follows:

$F_{wall} = \frac{\Delta p}{\Delta t} =\frac{(48.0 kg*(-1.06m/s)}{1.07s} = -47.6 N$

⇒ Fsk = 47.6 N (normal to the wall)

3 0

3 years ago

A weightlifter raises a 50kg weight to a height of 2m in 2 minutes. What was the power spent by the weightlifter?

belka [17]

Answer:

117.72kW

Explanation:

Given data

Mass m= 50kg

height x = 2m

time taken = 2 minutes= 129 seconds

let us find the work done

WD= force * distance

WD= mgx

WD= 50*9.81*2

WD= 981 Joules

Let us find the power

Power= work * time

Power= 981*120

Power= 117720

Power= 117.72 kW

Hence the power spent is 117.72kW

8 0

3 years ago

Use the following statements to determine if the disjunction is true or false. p: 7 is an odd number q: 3 is the square root of

kotykmax [81]

P and q are both true

3 0

2 years ago