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3 years ago

A student asks the following question:

Physics

1 answer:

MariettaO [177]3 years ago

8 0

Answer:

Natalie says that all things with mass have a gravitational field, but the force is very weak and cannot be perceived around small objects.

Explanation:

The force due to gravity is proportional to the mass of the object and inversely proportional to the square of the distance between objects. The Earth is so massive that the force due to its gravity is much greater than the force between objects on the counter.

If there were no friction, the objects might move toward each other, depending on what other masses were near them tending to cause them to move in other directions.

Natalie's explanation is about the best.

<em>Additional comment</em>

The universal gravitational constant was determined by Henry Cavendish in the late 18th century using lead balls weighing 1.6 pounds and 348 pounds. His experiment was enclosed in a large wooden box to minimize outside effects. While these masses are somewhat greater than those of a glue bottle and stapler, the experiment shows the force of gravity between "small" objects <em>can</em> be measured.

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A 1500 kg car traveling 5.0 m/s collides head on with a 3000 kg truck traveling

Setler [38]

m1=1500kg
m_2=3000kg
v_1=5m/s
v_2=7m/s

Using law of conservation of momentum

$\\ \sf\Rrightarrow m_1v_1-m_2v_2=(m1+m2)v_3$

$\\ \sf\Rrightarrow 1500(5)-3000(7)=(1500+3000)v_3$

$\\ \sf\Rrightarrow 7500-21000=4500v_3$

$\\ \sf\Rrightarrow -13500=4500v_3$

$\\ \sf\Rrightarrow v_3=-3m/s$

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2 years ago

Which of the following choices correctly identifies the process created by moving a magnet through a conducting loop?

jolli1 [7]

Answer: D.) electromagnetic induction

Explanation: Electroctromagnetic induction may be explained as a process whereby electric current is induced or produced by difference in potential resulting from the movement of conductor across a magnetic field.

In simple terms, an electromotive force is induced when a magnet is moved through a conducting loop.

The electromotive force produced by moving a magnet through a conducting loop can be represented by the relation:

E = - N (dΦ / dt)

Where E = electromotive force in voltage

N = number of loop in conductor

dΦ = change in magnetic Flux

dt = change in time

3 0

3 years ago

CAN SOMEONE PLEASE HELP ME!!!!! IM BEGGING YOU!!

murzikaleks [220]

In my opinion the answer is B. Variation

5 0

3 years ago

When the speed of the bottle is 2 m/s, the KE is kg m2/s2. When the speed of the bottle is 3 m/s, the KE is kg m2/s2. When the s

d1i1m1o1n [39]

mass of the bottle in each case is M = 0.250 kg

now as per given speeds we can use the formula of kinetic energy to find it

1) when speed is 2 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(2)^2 = 0.5 J$

2) when speed is 3 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(3)^2 = 1.125 J$

3) when speed is 4 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(4)^2 = 2 J$

4) when speed is 5 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(5)^2 = 3.125 J$

5) when speed is 6 m/s

kinetic energy is given as

$K = \frac{1}{2}mv^2$

$K = \frac{1}{2}(0.250)(6)^2 = 4.5 J$

3 0

3 years ago

Read 2 more answers

A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.

arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

Mass of the first puck = $m_1\ =\ 5\ kg$
Mass of the second puck = $m_2\ =\ 3\ kg$
initial velocity of the first puck = $u_1\ =\ 3\ m/s.$
Initial velocity of the second puck = $u_2\ =\ -1.5\ m/s.$

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.$

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = $v_2\ =\ 2.31\ m/s.$

Let $v_1$ be the final velocity of first puck after the collision.

From the conservation of linear momentum,

$m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.$

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

$v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\$

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = $25\times 10^{-3}\ sec$

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

$\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N$

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0

3 years ago