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2 years ago

Tell me what the

Physics

1 answer:

Talja [164]2 years ago

7 0

Answer:

upward thrust, makes the rocket go up

and kentic energy, you swing and the energy that was built up before you hit the ball was transferred

Explanation:

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A golf ball is hit with an initial velocity of 15 meters per second at an angle of 35 degrees above horizontal. What is the vert

polet [3.4K]

Answer:

8.6 m/s

Explanation:

The components of the initial velocity of the ball can be found by using the equations:

$v_x = v cos \theta\\v_y = v sin \theta$

where

$v_x$ is the horizontal component

$v_y$ is the vertical component

v is the magnitude of the velocity

$\theta$ is the angle above the horizontal

Substituting:

v = 15 m/s

$\theta=35^{\circ}$

We find:

$v_x = (15)(cos 35^{\circ})=12.3 m/s\\v_y = (15)(sin 35^{\circ})=8.6 m/s$

6 0

2 years ago

A spherical mirror is polished on both sides. When the concave side is used as a mirror, the magnification is 4.4. What is the m

patriot [66]

Answer:

$m = \frac{-4.4}{7.4}$

Explanation:

magnification is the ratio of distance of image to distance of object

i.e. $m = -\frac{d_i}{d_o}$

$d_i = -4.4 d_o$

As per the lens equation,

$\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}$

We will calculate the focal length of the mirror

$\frac{1}{yd_o} + \frac{1}{-4.4d_o} = \frac{1}{f} \\\frac{4.4 -1}{4.4} \frac{1}{d_o} = \frac{1}{f}\\f = \frac{4.4}{3.4} d_o$

Now for convex mirror only the sign will change

Thus, focal length would be equal to

$f = - \frac{4.4}{3.4} d_o$

Plugging this value into lens equation, we get

$\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f} \\\frac{1}{d_o} + \frac{1}{d_i} = \frac{-3.4}{4.4} \frac{1}{d_o} \\\frac{7.4}{4.4} \frac{1}{d_o} = \frac{1}{d_i}\\\frac{d_i}{d_o} = \frac{4.4}{7.4} \\m =- \frac{d_i}{d_o} \\\\m = - \frac{4.4}{7.4}$

8 0

3 years ago

alekssr [168]

Answer:

positive

proton and neutron

electron

7 0

2 years ago

PLEASE HELP URGENT!!!!!!!!!!!!!

Nimfa-mama [501]

c i think sorry if wrong

4 0

3 years ago

A small sphere has a harge of 9uC and other small sphere has a charge of 4uC.

Helga [31]

Answer:

Electrical force, F = 90 N

Explanation:

It is given that,

Charge on sphere 1, $q_1=9\ \mu C=9\times 10^{-6}\ C$

Charge on sphere 2, $q_1=4\ \mu C=4\times 10^{-6}\ C$

Distance between two spheres, d = 6 cm = 0.06 m

Let F is the electrical force between them. It is given by the formula of electric force which is directly proportional to the product of charges and inversely proportional to the square of distance between them such that,

$F=k\dfrac{q_1q_2}{d^2}$

$F=9\times 10^9\times \dfrac{9\times 10^{-6}\times 4\times 10^{-6}}{(0.06)^2}$

F = 90 N

So, the electrical force between them is 90 N. Hence, this is the required solution.

7 0

2 years ago