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3 years ago

What gives an object gravity?

1 answer:

8 0

We don't know WHY there are gravitational forces that attract everything
to everything else.

We DO know that the more mass is involved, the stronger the forces are,
and we know that if an object had no mass then it wouldn't be affected by
gravity. So you could say that it's MASS that gives an object gravity, and
the more mass the object has, the more gravity it has.

You might be interested in

You have 100 grams of different substances, the most dense substance would have which volume? A. 1L B. 0.10 L. C. 100 L D. 10 L

guajiro [1.7K]

Density is a value for mass, such as kg, divided by a value for volume, such as m3. Density is a physical property of a substance that represents the mass of that substance per unit volume. We do as follows:

A. 1L
100 g /1L = 100 g/L

B. 0.10 L.
100g/0.10L = 1000 g/L ------> MOST DENSE SUBSTANCE

C. 100 L
100g/100L = 1 g / L 

D. 10 L
100g/10L = 10 g/L

6 0

3 years ago

Block m1 of mass 2m and velocity v0 is traveling to the right (+x) and makes an elastic head-on collision with block m2 of mass

Oksana_A [137]

1) In any collision the momentum is conserved

(2*m)*(vo) + (m)*(-2*vo) = (2*m)(v1') + (m)(v2')

candel all the m factors (because they appear in all the terms on both sides of the equation)

2(vo) - 2(vo) = 2(v1') + (v2') => 2(v1') + v(2') = 0 => (v2') = - 2(v1')

2) Elastic collision => conservation of energy

=> [1/2] (2*m) (vo)^2 + [1/2](m)*(2*vo)^2 = [1/2](2*m)(v1')^2 + [1/2](m)(v2')^2

cancel all the 1/2 and m factors =>

2(vo)^2 + 4(vo)^2 = 2(v1')^2 + (v2')^2 =>

4(vo)^2 = 2(v1')^2 + (v2')^2

now replace (v2') = -2(v1')

=> 4(vo)^2 = 2(v1')^2 + [-2(v1')]^2 = 2(v1')^2 + 4(v1')^2 = 6(v1')^2 =>

(v1')^2 = [4/6] (vo)^2 =>

(v1')^2 = [2/3] (vo)^2 =>

(v1') = [√(2/3)]*(vo)

Answer: (v1') = [√(2/3)]*(vo)

4 0

3 years ago

Roger the turtle moved 450 m in 2 hours the first day and the next day he moved in a straight line at a steady speed of 15 cm/s

Sonja [21]

Answer:

a). Roger travel total a distance of 2070m

b). The average speed of the whole tip $V_{p}$ 0.16025 $\frac{m}{s}$

Explanation:

$V= \frac{x}{t} = \frac{450m}{2 hr} * \frac{1 hr}{3600 s} = \frac{450m}{7200 s}\\ V= \frac{450m}{7200s} =0.125 \\V= 0.0625 \frac{m}{s} \\$

a).

First trip 450m

Second trip :

$V=\frac{x}{t} \\x=V*t \\x= 15 \frac{cm}{s} *3 hr\\x= 15 (\frac{cm}{s}*\frac{1m}{100 cm}) * 3 hr (\frac{3600s}{1hr})\\ x=0.15 \frac{m}{s} * 3600s\\x= 1620 m$

So final distance is

450m+1620m=2070m

b).

$V_{p} = \frac{0.15 \frac{m}{s}+0.0625 \frac{m}{s} }{2} =0.10625 \frac{m}{s}$

The average speed is $V_{p}$ 0.10625 $\frac{m}{s}$

8 0

3 years ago

A 1000 Kg car approaches an intersection traveling north at 30 m/s . A 1250 Kg car approaches the same intersection traveling ea

Feliz [49]

Answer:

The final velocity has a magnitude of 25.44 m/s and is at 31.61° north of east.

Explanation:

Taking north direction as positive y axis and east direction as positive x axis .

Given:

Mass of first car is, $m_1=1000\ kg$

Initial velocity of first car is, $u_1=30\vec{j}\ m/s$

Mass of second car is, $m_2=1250\ kg$

Initial velocity of second car is,

Let the combined final velocity after collision be 'v' m/s with as components of final velocity along east and north directions respectively.

Now, as the net external force is zero, momentum is conserved for the two car system along the east and north directions.

Conserving momentum along the east direction, we have:

Initial momentum = Final momentum

$m_1u_{1x}+m_2u_{2x}=(m_1+m_2)v_x\\\\0+1250\times 39=(1000+1250)v_x\\\\v_x=\frac{48750}{2250}\\\\v_x=\frac{65}{3} m/s$

There is no component of initial velocity for first car in east direction, as it is moving in the north direction. So,

Now, conserving momentum along the north direction, we have:

Initial momentum = Final momentum

$m_1u_{1y}+m_2u_{2y}=(m_1+m_2)v_y\\\\1000\times 30+0=(1000+1250)v_y\\\\v_y=\frac{30000}{2250}\\\\v_y=\frac{40}{3}\ m/s$

There is no component of initial velocity for second car in north direction, as it is moving in the east direction. So, $u_{2y}=0$ .

The magnitude of final velocity is given as:

$|\vec{v}|=\sqrt{(v_x)^2+(v_y)^2}\\\\|\vec{v}|=\sqrt{(\frac{65}{3})^2+(\frac{40}{3})^2}\\\\|\vec{v}|=\sqrt{\frac{5825}{9}}=25.44\ m/s$

The direction is given as:

$\theta=\tan^{-1}(\frac{v_y}{v_x})\\\\\theta=\tan^{-1}(\frac{\frac{40}{3}}{\frac{65}{3}})=31.61^\circ$

So, the final velocity has a magnitude of 25.44 m/s and is at 31.61° north of east.

7 0

3 years ago

An object accelerates from 20 m/s down to 10 m/s in 5 seconds.

DedPeter [7]

Answer:

The acceleration of the object is $-2\ \frac{m}{sec^2}$

Explanation:

we know that

The acceleration is the change in the velocity, divided by the time

Let

v -----> the change in the velocity in m/sec

t ----> the time in sec

$a=\frac{V}{t}$

we have that

$V_2=10\ m/sec\\V_1=20\ m/sec$

$V=V_2-V_1$

substitute

$V=10-20=-10\ m/sec$

$t=5\ sec$

substitute in the formula

$a=\frac{-10}{5}=-2\ \frac{m}{sec^2}$

therefore

The acceleration of the object is $-2\ \frac{m}{sec^2}$

Is negative because is a deceleration

5 0

3 years ago