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3 years ago

What TWO things affect the gravitational force acting on an object?

2 answers:

5 0

<u><em>mass and distance</em></u> (between two objects) are those TWO things that affect the gravitational force acting on an object

ohaa [14]3 years ago

3 0

The strength of the gravitational forces between two masses depends on

-- the product of the masses,

-- the distance between their centers of mass.

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Suppose the value of one division of vernier scale is 0.5mm and the value of one main scale

gregori [183]

Answer:

-0.01 mm

Explanation:

We are given that

The value of one division of vernier scale =0.5 mm

The value of one main scale division=0.49 mm

We have to find the value of least count of the instrument in mm.

We know that

Leas count of vernier caliper=1 main scale division-1 vernier scale division

Least count of vernier caliper=0.49-0.50=-0.01 mm

Hence, the least count of the instrument=-0.01 mm

Answer: -0.01 mm

8 0

3 years ago

What is the value of acceleration due to gravity at the pole, equator and the centre of the earth

fgiga [73]

Answer:

In combination, the equatorial bulge and the effects of the surface centrifugal force due to rotation mean that sea-level gravity increases from about 9.780 m/s2 at the Equator to about 9.832 m/s2 at the poles, so an object will weigh approximately 0.5% more at the poles than at the Equator.

7 0

2 years ago

A convex lens has a focal length of 16.5 cm. Where on the lens axis should an object be placed in order to get a virtual, enlarg

alexandr402 [8]

Answer:

Object should be placed at a distance, u = 7.8 cm

Given:

focal length of convex lens, F = 16.5 cm

magnification, m = 1.90

Solution:

Magnification of lens, m = - $\frac{v}{u}$

where

u = object distance

v = image distance

Now,

1.90 = $\frac{v}{u}$

v = - 1.90u

To calculate the object distance, u by lens maker formula given by:

$\frac{1}{F} = \frac{1}{u}+ \frac{1}{v}$

$\frac{1}{16.5} = \frac{1}{u}+ \frac{1}{- 1.90u}$

$\frac{1}{16.5} = \frac{1.90 - 1}{1.90u}$

$\frac{1}{16.5} = \frac{ 0.90}{1.90u}$

u = 7.8 cm

Object should be placed at a distance of 7.8 cm on the axis of the lens to get virtual and enlarged image.

6 0

4 years ago

SYW A force of 175 N is needed to keep a stationary engine of weight 640 N on wooden skids from

grigory [225]

Answer:

phle follow karo yrr tab hee bat u ga

4 0

3 years ago

Read 2 more answers

Arocket launches at an angle of 33.6 degrees from the horizontal at a

babymother [125]

Answer:

Y component = 32.37

Explanation:

Given:

Angle of projection of the rocket is, $\theta=33.6$

Initial velocity of the rocket is, $u=58.5$

A vector at an angle $\theta$ with the horizontal can be resolved into mutually perpendicular components; one along the horizontal direction and the other along the vertical direction.

If a vector 'A' makes angle $\theta$ with the horizontal, then the horizontal and vertical components are given as:

$A_x=A\cos \theta(\textrm{Horizontal or X component})\\A_y=A\sin \theta(\textrm{Vertical or Y component})$

Here, as the velocity is a vector quantity and makes an angle of 33.6 with the horizontal, its Y component is given as:

$u_y=u\sin \theta$

Plug in the given values and solve for $u_y$ . This gives,

$u_y=(58.5)(\sin 33.6)\\u_y=58.5\times 0.55339\\u_y=32.373\approx32.37(\textrm{Rounded to two decimal places})$

Therefore, the Y component of initial velocity is 32.37.

4 0

3 years ago