All categories

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.

You might be interested in

Charge is uniformly distributed around a ring of radius R and the resulting electric field magnitude E is measured along the rin

Arte-miy333 [17]

Answer:

$x=\dfrac{r}{\sqrt2}$

Explanation:

Given that

Radius =r

Electric filed =E

Q=Charge on the ring

The electric filed at distance x given as

$E=K\dfrac{Q}{(r^2+x^2)^{3/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$E=K{Q}{(r^2+x^2)^{-3/2}}$

$\dfrac{dE}{dx}=K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}$

For maximum condition

$\dfrac{dE}{dx}=0$

$K{Q}{(r^2+x^2)^{-3/2}}-\dfrac{3}{2}\times 2\times x\times K{Q}{(r^2+x^2)^{-5/2}}=0$

$r^2+x^2-3x^2=0$

$x=\dfrac{r}{\sqrt2}$

At $x=\dfrac{r}{\sqrt2}$ the electric field will be maximum.

3 0

2 years ago

Long Jump: inital center of mass height of 1.08 m, final center of mass height of 0.42 m, projection velocity of 8.7 m/s, projec

sammy [17]

Answer:

1) The maximum jump height is reached at A. $0.337s$

2) The maximum center of mass height off of the ground is B. $1.64m$

3) The time of flight is C. $0.834s$

4) The distance of jump is B. $7.49m$

Explanation:

First of all we need to decompose velocity in its rectangular components, so

$v_{xi}=8.7m/s(cos 22.3\°)=8.05m/s= constant\\v_{yi}=8.7m/s(sin 22.3\°)=3.3m/s$

1) We use, $v_{fy}=v_{iy}-gt$ , as we clear it for $t$ and using the fact that $v_{fy}=0$ at max height, we obtain $t=\frac{v_{iy}}{g} =\frac{3.3m/s}{9,8m/s^{2}} =0.337s$

2) We can use the formula $y_{max}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ for $t=0.337s$ , so

$y_{max}=1.08m+(3.3m/s)(0.337s)-\frac{(9.8m/s^{2})(0.337)^{2}}{2}=1.64m$

3) We can use the formula $y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}$ , to find total time of fligth, so $0.42=1.08+3.3t-\frac{(9.8)t^{2}}{2}\\0=-4.9t^{2}+3.3t+0.66$ , as it is a second-grade polynomial, we find that its positive root is $t=0.834s$

4) Finally, we use $x=v_{x}t=8.05m/s(0.834s)=6.71m$ , as it has an additional displacement of $0.77m$ due the leg extension we obtain,

$x=6.71m+0.77m=7.48m$ , aprox $x=7.49m$

5 0

3 years ago

With the aim of a diagram describe how a rain bow is formed

Natalija [7]

Answer:

Water droplets acts as tiny prism in the sky. The sunlight when enters these tiny droplets undergo internal reflection and also refract these rays which are dispersed causing a band of seven colors called rainbow.

Explanation:

3 0

2 years ago

Read 2 more answers

A soccer ball with mass 0.450 kg is initially moving with speed 2.20 m/s. A soccer player kicks the ball, exerting a constant fo

Alinara [238K]

Answer:

0.187 m

Explanation:

We'll begin by calculating the acceleration of the ball. This can be obtained as follow:

Mass (m) = 0.450 Kg

Force (F) = 38 N

Acceleration (a) =?

F = m × a

38 = 0.450 × a

Divide both side by 0.450

a = 38 / 0.450

a = 84.44 m/s²

Finally, we shall determine the distance. This can be obtained as follow:

Initial velocity (u) = 2.20 m/s.

Final velocity (v) = 6 m/s

Acceleration (a) = 84.44 m/s²

Distance (s) =?

v² = u² + 2as

6² = 2.2² + (2 × 84.44 × s)

36 = 4.4 + 168.88s

Collect like terms

36 – 4.84 = 168.88s

31.52 = 168.88s

Divide both side by 168.88

s = 31.52 / 168.88

s = 0.187 m

Thus, the distance is 0.187 m

6 0

3 years ago

What is the kinetic energy of a 0.135 kg baseball thrown at 40 m/s

Ronch [10]

KE = 1/2 * m * v^2
KE = 1/2 * 0.135 * 40^2
KE = 1/2 * 0.135 * 1600
KE = 108 J

4 0

3 years ago