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3 years ago

8

A ball is thrown vertically upwards with a velocity of 20 m s −1 from the top of a multistorey building of 25 m high. How high w

ill the ball rise from top of the building? (Take g = 10 m s

1 answer:

Sunny_sXe [5.5K]3 years ago

3 0

Answer:

20m

Explanation:

How high the ball rise from top of the building is the same as the maximum height reached by the ball

using equation of motion under gravity

V^2=U^2+2gs

making s the subject

U^2/2g

20^2/2×10=20m

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Consider a car that travels between points A and B. The car's average speed can be greater than the magnitude of its average vel

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3 years ago

Melting butter is a physical change. Which best describes what is happening?

kati45 [8]

Answer:

Option B

Explanation:

The butter forms a new arrangement of atoms i.e from solid to liquid

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How does mass affect weight?

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Answer: Mass affects the weight of an object with the effects of gravity.

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3 years ago

Someone please help with this

SSSSS [86.1K]

Answer:

<em>The new force is 2/3 of the original force</em>

Explanation:

<u>Coulomb's Law </u>

The electrical force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between the two objects.

Written as a formula:

$\displaystyle F=k\frac{q_1q_2}{d^2}$

Where:

$k=9\cdot 10^9\ N.m^2/c^2$

q1, q2 = the objects' charge

d= The distance between the objects

Suppose the first charge is doubled (2q1) and the second charge is one-third of the original charge (q2/3). Now the force is:

$\displaystyle F'=k\frac{2q_1*q_2/3}{d^2}$

Factoring out 2/3:

$\displaystyle F'=\frac{2}{3}k\frac{q_1*q_2}{d^2}$

Substituting the original force:

$F'=\frac{2}{3}F$

The new force is 2/3 of the original force

7 0

3 years ago

A 0.29 kg particle moves in an xy plane according to x(t) = - 19 + 1 t - 3 t3 and y(t) = 20 + 7 t - 9 t2, with x and y in meters

Artist 52 [7]

Answer:

Part a)

$F = 7.76 N$

Part b)

$\theta = -137.7 degree$

Part c)

$\theta = -127.7 degree$

Explanation:

As we know that acceleration is rate of change in velocity of the object

So here we know that

$x = -19 + t - 3t^3$

$y = 20 + 7t - 9t^2$

Part a)

differentiate x and y two times with respect to time to find the acceleration

$a_x = \frac{d^2}{dt^2}(-19 + t - 3t^3)$

$a_x = \frac{d}{dt}(0 +1 - 9t^2)$

$a_x = -18t$

$a_y = \frac{d^2}{dt^2}(20 + 7t - 9t^2)$

$a_y = \frac{d}{dt}(0 +7 - 18t)$

$a_y = -18$

Now the acceleration of the object is given as

$\vec a = (-18t)\hat i + (-18)\hat j$

at t= 1.1 s we have

$\vec a = -19.8 \hat i - 18 \hat j$

now the net force of the object is given as

$\vec F = m\vec a$

$\vec F = (0.29 kg)(-19.8 \hat i - 18 \hat j)$

$\vec F = -5.74 \hat i - 5.22 \hat j$

now magnitude of the force will be

$F = \sqrt{5.74^2 + 5.22^2} = 7.76 N$

Part b)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-5.22}{-5.74}$

$\theta = -137.7 degree$

Part c)

For velocity of the particle we have

$v_x = \frac{dx}[dt}$

$v_x = (0 +1 - 9t^2)$

$v_y = \frac{dy}{dt}$

$v_y = (0 +7 - 18t)$

now at t = 1.1 s

$\vec v = -9.89\hat i - 12.8 \hat j$

now the direction of the velocity is given as

$\theta = tan^{-1}(\frac{v_y}{v_x})$

$\theta = tan^{-1}(\frac{-12.8}{-9.89})$

$\theta = -127.7 degree$

7 0

3 years ago