Ask question

Ask question

All categories

antoniya [11.8K]

2 years ago

13

Johanna is studying what happens to the energy as a ball rolls down a ramp. What is one form of energy that she is studying? gra

vitational potential energy electrical energy chemical energy elastic potential energy

2 answers:

Rudiy272 years ago

4 0

Answer:

the answer is A

Explanation:

gravitational potential energy

timofeeve [1]2 years ago

3 0

Answer: gravitational potential energy

Explanation:

Gravitational potential energy is refered to as the potential energy that an object has due to its gravitational field, which in turn is being turned into a kinetic energy when there is a fall of such objects.

Since Johanna is studying what happens to the energy as a ball rolls down a ramp, the form of energy here is the gravitational potential energy.

You might be interested in

Boyle’s Law states that when a sample of gas is compressed at a constant temperature, the pressure P of the gas is inversely pro

jolli1 [7]

Answer:

a) $V=\dfrac{5.3}{P}$

b) $ML^{-4}T^{-2}$ .

Explanation:

Given that

Boyle's law

P V = Constant ,at constant temperature

a)

Given that

$P_1=50KPa$

$V_1=0.106m^3$

We know that for PV=C

$P_1V_1=P_2V_2=PV$

Now by putting the values

PV= 50 x 0.106

$V=\dfrac{5.3}{P}$

Where P is in KPa and V is in $m^3$

b)

PV= C

Take ln both sides

So $\ln(PV)=\ln C$

lnP + lnV =lnC ( C is constant)

By differentiating

$\dfrac{dP}{P}+\dfrac{dV}{V}=0$

So

$\dfrac{dP}{dV}=-\dfrac{P}{V}$

When P= 50 KPa

$\dfrac{dP}{dV}=-\dfrac{50}{V}\ \dfrac{KPa}{m^3}$

It indicates the slope of PV=C curve.

It unit is $\dfrac{Pa}{m^3}$ .

Or we can say that $ML^{-4}T^{-2}$ .

5 0

3 years ago

Read 2 more answers

How much work did the movers do (horizontally) pushing a 41.0-kg crate 10.6 m across a rough floor without acceleration, if the

VLD [36.1K]

Answer:

The required work done is $2555.448~J$

Explanation:

Consider 'F' is the applied force on the crate and 'f' be the force created by friction. According to the figure if ' $\mu_{k}$ ' be the coefficient of friction, then

$f = \mu_{k} \times N = \mu_{k} \times Mg$

where 'M', 'N' and 'g' are the mass of the crate, the normal force aced upon the block and the acceleration due to gravity respectively.

Since the application of force by the movers does not create any acceleration to the block, we can write

$F = f = \mu_{k} \times M \times g = 0.6 \times 41~Kg~ \times 9.8~m~s^{-2} = 241.08~N$

So the work done (W) in moving the crate by a distance s = 10.6 m is

$W = F \times s = 241.08~N \times 10.6~m = 2555.448 J$

5 0

3 years ago

What is the magnitude of the acceleration of a speck of clay on the edge of a potter's wheel turning at 45 rpm (revolutions per

Brilliant_brown [7]

Answer:

$a_{cp}=7.77m/s^2$

Explanation:

The equation for centripetal acceleration is $a_{cp}=\frac{v^2}{r}$ .

We know the wheel turns at 45 rpm, which means 0.75 revolutions per second (dividing by 60), so our frequency is f=0.75Hz, which is the inverse of the period T.

Our velocity is the relation between the distance traveled and the time taken, so is the relation between the circumference $C=2\pi r$ and the period T, then we have:

$v=\frac{C}{T}=2\pi r f$

Putting all together:

$a_{cp}=\frac{(2\pi r f)^2}{r}=4 \pi^2f^2r=4 \pi^2(0.75Hz)^2(0.35m)=7.77m/s^2$

4 0

3 years ago

The density of table sugar is 1.59 g/cm3. What is the volume of 7.85 g of sugar?

Likurg_2 [28]

$\frac{7.85g }{1.59 \frac{g}{ cm^{3} } }$ = 4.937 $cm^{3}$

4 0

3 years ago

mezya [45]

The advantage is that we do not run out of resources and a disadvantage is that is dangerous when a “human” gets too close and gets sick by the radiation.

5 0

2 years ago