How much net force is needed to accelerate a 15 kg mass at 2.8 m/s^2

A lamp consumes 1000J of ekectrical energy in 10s. Calculate its power.

Mice21 [21]

You do 1000 divide it by 10 which equals 100 W

4 0

3 years ago

A mass M is attached to an ideal massless spring. When this system is set in motion with amplitude A, it has a period T. What is

cricket20 [7]

Answer:

The period of the motion will still be equal to T.

Explanation:

for a system with mass = M

attached to a massless spring.

If the system is set in motion with an amplitude (distance from equilibrium position) A

and has period T

The equation for the period T is given as

$T = 2\pi \sqrt{\frac{M}{k} }$

where k is the spring constant

If the amplitude is doubled, the distance from equilibrium position to the displacement is doubled.

Increasing the amplitude also increases the restoring force. An increase in the restoring force means the mass is now accelerated to cover more distance in the same period, so the restoring force cancels the effect of the increase in amplitude. Hence, increasing the amplitude has no effect on the period of the mass and spring system.

5 0

3 years ago

You’ve made the finals of the science Olympics. As one of your tasks you’re given 1.0 g of copper and asked to make a cylindrica

Pani-rosa [81]

Answer:

Length = 2.92 m

Diameter = 0.11 mm

Explanation:

We have $m = dl D \ \ \& \ \ \ R = \frac{\rho l}{A}$ , where:

$l$ is the length

$m = 1.0 g = 1 \times 10^{-3} \ kg\\R = 1.3 \ \Omega\\\rho = 1.7 \times 10^{-8} \Omega m\\d = 8.96 \ g/cm^3 = 8960 kg/m^3$

We divide the first equation by the second equation to get:

$\frac{m}{R} = \frac{d A^2}{\rho}$

$A^2 = \frac{m \rho}{dR} \\\\A^2 = \frac { 1 \times 10^{-3} \times 1.7 \times 10^{-8}}{8960 \times 1.3}\\\\A^2 = 1.5 \times 10^{-15}\\\\ A= 3.8 \times 10^{-8} \ m^2$

Using this Area, we find the diameter of the wire:

$D = \sqrt{\frac{4A}{\pi}}$

$D = \sqrt{\frac{4 \times 3.8 \times 10^{-8} }{\pi}}$

$D = 0.00011 \ m = 1.1 \times 10^ {-4} = 0.11 \ mm$

To find the length, we multiply the two equations stated initially:

$mR = d\rho l^2\\\\l^2 = \frac{mR}{d\rho} \\\l^2 = \frac {1.0 \times 10^{-3} \times 1.3}{8960 \times 1.7\times 10^{-8}}$

$l^2 = 8.534\\l = 2.92 \ m$

8 0

3 years ago

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the force that is exerted when a shopping cart is pushed. the forces that causes a metal ball to move toward a magnet

alexandr402 [8]

Answer:

The force that is exerted when a shopping cart is pushed is a type of push force, supplied by the muscles of the cart pusher's body.

The forces that causes a metal ball to move toward a magnet is a type of pull force that is as a result of the magnetic field forces.

Explanation:

Forces are divided into push forces that tends to accelerate a body away from the source of the force, and pull forces that accelerates the body towards the force source.

Examples of push forces includes pushing a cart, pushing a table, repulsion of two similar poles of a magnet etc. Examples of pull forces includes a attractive force between two dissimilar poles of a magnet, pulling a load by a rope, a dog pulling on a leash etc.

8 0

3 years ago

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Based on the principles of convection, conduction and thermal radiation, which scenario below is most similar to the following s

Ksivusya [100]

The situation (heat going through the ceiling) describes
conduction ... heat going from one place to another by
soaking through some material.

A). This is the one. Heat goes from from the marshmallow
to your hand by soaking through the wire. This is conduction too.

B). No. The heat in the room goes from the floor to the ceiling
because the warm air rises and carries it there. This is convection.

C). No. There's nothing for the heat to soak through between
the sun and the roof, and nothing that can move from the sun
to the roof and bring the heat with it. This is radiation.

D). No. Cold water sinks from the surface to the bottom because
warm water rose from the bottom to the surface, taking heat with it.
This is convection.

5 0

3 years ago

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