Ask question

Ask question

All categories

3 years ago

14

During a contest that involved throwing a 7.0-kg bowling ball straight up in the air, one contestant exerted a force of 810 N on

the ball. If the force was exerted through a distance of 2.0 m, how high did the ball go from the point of release?

1 answer:

liubo4ka [24]3 years ago

5 0

Answer:

23.6 m

Explanation:

We are given that

Mass of ball, m=7.0 kg

Force, F=810 N

Distance, s=2.0 m

We have to find the height of ball from the point where it releases.

Work done=K.E

$Fs=\frac{1}{2}mv^2$

$v^2=\frac{2Fs}{m}$

Substitute the value

$v^2=\frac{2\times 810\times 2}{7}$

$v^2=\frac{3240}{7}$

K.E=P.E

$\frac{1}{2}mv^2=mgh$

$\frac{1}{2}v^2=gh$

Where g= $9.8m/s^2$

$\frac{1}{2}\times \frac{3240}{7}=9.8h$

$h=\frac{3240}{7\times 2\times 9.8}$

$h=23.6 m$

You might be interested in

How do low energy electromagnetic waves compare with high energy electromagnetic waves? Select all

Dennis_Churaev [7]

Answer:

Explanation:

The different types of radiation are defined by the the amount of energy found in the photons. Radio waves have photons with low energies, microwave photons have a little more energy than radio waves, infrared photons have still more, then visible, ultraviolet, X-rays, and, the most energetic of all, gamma-rays

3 0

3 years ago

A robotic rover on Mars finds a spherical rock with a diameter of 10 centimeters [cm]. The rover picks up the rock and lifts it

Makovka662 [10]

Answer: 5166.347

Explanation:

The specific gravity of a solid $SG$ (also called relative density) is the ratio of the density of that solid $\rho_{rock}$ to the density of water $\rho_{water}=1 kg/m^{3}$ (normally at $4\°C$ ):

$SG=\frac{\rho_{rock}}{\rho_{water}}$ (1)

On the other hand, the density of the rock is calculated by:

$\rho_{rock}=\frac{m_{rock}}{V_{rock}}$ (2)

Where:

$m_{rock}$ is the mass of the rock

$V_{rock}=\frac{4}{3} \pi r^{3}$ is the volume of the rock, since is spherical

Well, we already know the value of $\rho_{water}$ , but we need to find $\rho_{rock}$ in order to find the rock's specific gravity; and in order to do this, we firsly have to find $m_{rock}$ and then calculate $V_{rock}$ :

In the case of the mass of the rock, we can calclate it by the following equation:

$W_{rock}=m_{rock}g_{mars}$ (3)

Where:

$W_{rock}$ is the weight if the rock in mars

$g_{mars}=3.7 m/s^{2}$ is the acceleration due gravity in Mars

Isolating $m_{rock}$ :

$m_{rock}=\frac{W_{rock}}{g_{mars}}$ (4)

$m_{rock}=\frac{W_{rock}}{3.7 m/s^{2}}$ (5)

To find $W_{rock}$ we can use the following equation of the potential gravitational energy $U$ :

$U=W_{rock}H$ (6)

Where:

$U=2 J=2 Nm$ is the potential energy

$H=20 cm \frac{1m}{100 cm}=0.2 m$ is the height at which the rock has the mentioned potential energy

Isolating $W_{rock}$ :

$W_{rock}=\frac{U}{H}$ (7)

$W_{rock}=\frac{2 Nm}{0.2 m}$ (8)

$W_{rock}=10 N$ (9)

Substituting (9) in (5):

$m_{rock}=\frac{10 N}{3.7 m/s^{2}}$ (10)

$m_{rock}=2.702 kg$ (11)

Substituting (11) in (2):

$\rho_{rock}=\frac{2.702 kg}{V_{rock}}$ (12) At this point we only need to find the volume of the rock, knowing its diameter is $d=10 cm$ , hence its radius is $r=\frac{d}{2}=5 cm$

$V_{rock}=\frac{4}{3} \pi (5 cm)^{3}$ (13)

$V_{rock}=523.59 cm^{3} \frac{1 m^{3}}{(100 cm)^{3}}=0.000523 m^{3}$ (14)

Substituting (14) in (12):

$\rho_{rock}=\frac{2.702 kg}{0.000523 m^{3}}$ (15)

$\rho_{rock}=5166.34 kg/m^{3}$ (16)

Substituting (16) in (1):

$SG=\frac{5166.34 kg/m^{3}}{1 kg/m^{3}}$ (17)

Finally we obtain the specific gravity of the rock:

$SG=5166.347$

7 0

3 years ago

How are physical traits inherited from one generation to the next?

Umnica [9.8K]

Http://www.nature.com/scitable/topicpage/inheritance-of-traits-by-offspring-follows-predictable-6524... this should answer all questions you have on this specific subject in science

3 0

4 years ago

What happens to the amount of energy present when it is moved from one form to another

Jet001 [13]

The amount of energy is ALWAYS conserved based on the law of conservation of energy. Therefore, the amount of energy will remain the same.

3 0

3 years ago

A geothermal plant uses geothermal water (yes a liquid) extracted from beneath the Earth using drilling apparatus. The geotherma

leva [86]

To solve this problem it is necessary to apply the concepts of heat change and thermal efficiency.

The heat rate can be expressed under the function

$Q = \dot{m} (h_1-h_2)$

Where,

m = Mass

$h_i =$ Enthalpy at each state

Our values are given as,

$\dot{m} = 200kg/s$

$T_H = 200\°C$

$W = 6000kW$

$T_{H,2} = 90\°C$

$T_L = 25\°C$

From the tables of Enthalpy of Water at 200°C (Saturated liquid state)

$h_1 = 852.4kJ/Kg$

At the same time for 80°C

$h_2 = 334.9kJ/Kg$

Applying the equation of Heat,

$Q = \dot{m}(h_1-h_2)$

Replacing,

$Q = 200*(852.4-334.9)$

$Q = 103500kW$

Therefore the efficiency would be

$\eta = \frac{Q_L}{Q_H}$

$\eta = \frac{6000}{103500}$

$\eta = 0.0579$

Therefore the actual thermal efficiency of the turbine in percent is 0.0579.

6 0

3 years ago