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3 years ago

Q. Two balls are released simultaneously from a certain

Physics

1 answer:

Mice21 [21]3 years ago

5 0

Answer:Yes

Explanation:

Time taken for the first ball will depend upon the height of roof. As it is dropped so initial vertical velocity will be zero.

so time taken will be

$t=\sqrt{\frac{2h}{g}}$

where h=height of roof

g=acceleration due to gravity

Horizontal velocity is given to second ball so it cover more horizontal distance as compared to first but time taken will be same as it has to cover same vertical distance with zero vertical velocity and time taken depends on height of roof and g.

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what is the kinetic energy of an object that has mass of 30 kilograms and move with a velocity of 20 m/s

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Kinetic Energy = 1/2 * m * v² 1/2 * 30 * 20² 1/2 * 30 * 400 12000/2 6000 J.

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3 years ago

What is a force ?????

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Answer:

A force is a push or pull upon an object resulting from the objects interaction with another object

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3 years ago

Read 2 more answers

Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th

aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

3 0

3 years ago

What mass of a material with density rho is required to make a hollow spherical shell having inner radius r1 and outer radius r2

Margarita [4]

Answer:

$m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3 )$

Explanation:

We have to make a hollow sphere of inner radius $r_1$ and outer radius $r_2$ .

Then the mass of the material required to make such a sphere would be calculated as:

Total volume of the spherical shell:

$V_t=\frac{4}{3} \pi.r_2^3$

And the volume of the hollow space in the sphere:

$V_h=\frac{4}{3} \pi.r_1^3$

Therefore the net volume of material required to make the sphere:

$V=V_t-V_h$

$V=\frac{4}{3} \pi(r_2^3-r_1^3)$

Now let the density of the of the material be $\rho$ .

<u>Then the mass of the material used is:</u>

$m=\rho.V$

$m=\rho\times \frac{4}{3} \times \pi \times(r_2^3-r_1^3 )$

4 0

3 years ago

A balloon is filled with helium at sea level. The volume was measured to be 3.1 liters at a pressure of 0.97 atm. If the balloon

stira [4]

Answer:

4.9L hope this helped!

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3 years ago