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eimsori [14]
3 years ago
5

A 6 kg bowling ball moves with a speed of 3 m/s. How fast does a 7 kg bowling ball need to move so that it has the same kinetic

energy?
Physics
1 answer:
maw [93]3 years ago
4 0

Answer: 7 kg bowling ball must move with a speed of 2.8 m/s so that it has the same kinetic energy.

Explanation:

Kinetic energy is the energy possessed by a body by virtue of its motion.

K.E=\frac{1}{2}mv^2

m = mass of object

v= velocity of the object

K.E=\frac{1}{2}\times 6kg\times (3m/s)^2=27Joules

b) for a 7 kg bowl to have kinetic energy of 27 Joules:

27J=\frac{1}{2}\times 7kg\times v^2

v^2=7.7

v=2.8m/s

Thus 7 kg bowling ball must move with a speed of 2.8 m/s so that it has the same kinetic energy

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The maximum pressure most organisms can survive is about 1000 times the atmospheric pressure. Only small, simple organisms such
Schach [20]

Answer:

h = 10000 m

Explanation:

The pressure applied at a depth of the liquid is given by:

P =ρgh

where,

P = Maximum Pressure to Survive = (1000)(Atmospheric Pressure)

P = (1000)(101325 Pa) = 1.01 x 10⁸ Pa

ρ = Density of sea water = 1025 kg/m³

g = 9.8 m/s²

h = maximum depth to survive = ?

Therefore,

1.01 x 10⁸ Pa = (1025 kg/m³)(9.8 m/s²)h

h = (1.01 x 10⁸ Pa)/(1025 kg/m³)(9.8 m/s²)

<u>h = 10000 m</u>

6 0
3 years ago
if a ball is thrown straight up into the air with an initial velocity of 8080 ft/s, its height in feet after tt second is given
yanalaym [24]

The average velocity for the time period beginning when t=1 and lasting

(i) 0.01 seconds = 63.84 ft/s

(ii) 0.001 seconds = 63.984 ft/s

Given that a ball is thrown with an initial velocity = 80 ft/s

Let 'y' be the height in feet after 't' seconds.

Given,  y=80t-16t^2 gives the height in 't' seconds.

Average velocity = Rate of change of distance

                             = Change in distance/Change in time.

The initial time can be taken as 0 s.

When t =1 s, y = 80 - 16 = 64 ft

(1)  t = 0.01 s

    y = 80 x 0.01 - 16 x 0.01 x 0.01 = 0.7984 ft

    Average velocity = (64 - 0.7984) / (1 -0.01) = 63.84 ft/s

(2) t = 0.001 s

    y = 80 x 0.001 - 16 x 0.001 x 0.001 = 0.079984 ft

    Average velocity = (64 - 0.079984) / (1 -0.001) = 63.984 ft/s

The question is incomplete. Find out the complete question below:

If a ball is thrown straight up into the air with an initial velocity of 80 ft/s, it height in feet after t second is given by  y=80t-16t^2 .Find the average velocity for the time period beginning when t=1 and lasting

(i) 0.01 seconds

(ii) 0.001 seconds

Learn more about average velocity at brainly.com/question/6504879

#SPJ4

7 0
2 years ago
Which of the following would have the greatest force of gravitational attraction?
Mekhanik [1.2K]

Answer:

two 5kg objects 2 meters apart✨

8 0
3 years ago
An AM radio transmitter broadcasts 63.2 kW of power uniformly in all directions. (a) Assuming all of the radio waves that strike
egoroff_w [7]

Answer:

Intensity of the transmitted radio wave is 5.406 x 10⁻⁶ W/m²

Explanation:

Given;

power of radio transmitter, P = 63.2 kW = 63200 W

distance of transmission, r = 30.5 km

Intensity of the transmitted radio wave is calculated as follows;

I = \frac{P}{4\pi r^2}

where;

I is the intensity of the transmitted radio wave

Substitute the given values and calculate the intensity of the transmitted radio wave;

I = \frac{P}{4\pi r^2} = \frac{63200}{4\pi (30500)^2} = 5.406 *10^{-6} \ W/m^2

Therefore, Intensity of the transmitted radio wave is 5.406 x 10⁻⁶ W/m²

6 0
3 years ago
Josie ran at an average speed of 16 m/s. if her mass 10 kg, what was her kinetic energy as she crosses her finish line?
Helen [10]

Answer: 1280 J

KE=\frac{1}{2}mv^{2}  \\\\KE= \frac{1}{2} (10 kg)(16 m/s)^2\\=1280 J

3 0
2 years ago
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