Ask question

Ask question

All categories

3 years ago

7

A box is at rest at the top of a frictionless inclined plane. As the box slides down the ramp where along the ramp would the box

have half of its maximum kinetic energy?

1 answer:

Yuliya22 [10]3 years ago

3 0

Answer:

I would say near the end of the slide.

You might be interested in

How is it possible for one electrically neutral atom to exert an electrostatic force on other electrically neutral atom?

Free_Kalibri [48]

Answer:

Explained

Explanation:

Although Atom are electrically neutral. But atom atom is combination of nucleus and electrons. The nucleus of the atom is composed neutron and positively charged protons. On the outside of nucleus at some distance are the electrons which are negatively charged. So, there is difference in position of the two differently charge species. So, this way a electrically neutral atom can exert a electrostatic force on other electrically neutral atom

8 0

4 years ago

A stonecutter's chisel has an edge area of 0.7 cm2. If the chisel is struck with a force of 42 N, what is the pressure exerted o

Fantom [35]

Answer:

The pressure is $P = 583333 \ N/m^2$

Explanation:

From the question we are told that

The area of the edge is $A = 0.72 cm^2 = 0.72 *10^{-4}\ m$

The force is $F = 42 \ N$

The pressure is mathematically represented as

$P = \frac{F}{A}$

substituting values

$P = \frac{42}{0.72*10^{-4}}$

$P = 583333 \ N/m^2$

5 0

3 years ago

Presuming you have a closed circuit, when you thrust a bar magnet to and fro into a coil of wire, you induce:________.

Art [367]

Hedndjf Hinduism’s jeirbduddn dudurnidkd idbehdududid idjdjdudud idjdbdudud idisjdbdbd

7 0

3 years ago

A 450.0 N, uniform, 1.50 m bar is suspended horizontally by two vertical cables at each end. Cable A can support a maximum tensi

bogdanovich [222]

Answer:

1) $W_{object} = 400 N$

2) x = 0.28 m from cable A.

Explanation:

1 ) Let's use the first Newton to find the , because bar is in equilibrium.

$\sum F_{Tot} = 0$

In this case we just have y-direction forces.

$\sum F_{Tot} = T_{A}+T_{B}-W_{bar}-W_{object} = 0$

Now, let's solve the equation for W(object).

$W_{object} = T_{A}+T_{B}-W_{bar} = 550 +300 - 450 = 400 N$

2 ) To find the position of the heaviest weight we need to use the torque definition.

$\sum \tau = 0$

The total torque is evaluated in the axes of the object.

Let's put the heaviest weight in a x distance from the cable A. We will call this point P for instance.

First let's find the positions from each force to the P point.

$L = 1.50 m$ ; total length of the bar.

$D_{AP} = x$ ; distance between Tension A and P point.

$D_{BP} = L-x$ ; distance between Tension B and P point.

$D_{W_{bar}P} = \frac{L}{2}-x$ ; distance between weight of the bar (middle of the bar) and P point.

Now, let's find the total torque in P point, assuming counterclockwise rotation as positive.

$\sum \tau = T_{B}(L-x)-T_{A}(x)-W_{bar}(\frac{L}{2}-x) = 0$

Finally we just need to solve it for x.

$x = \frac{T_{B}L-W_{bar}(L/2)}{T_{B}+W_{bar}+T_{A}}$

$x = 0.28 m$

So the distance is x = 0.28 m from cable A.

Hope it helps!

Have a nice day! :)

8 0

4 years ago

An object has a mass of 50.0 g and a volume of 10.5 cm3. What is the object's density?

givi [52]

Divide the objects mass by its volume. 50.0g / 10.5cm^3 =4.76cm ^3

8 0

3 years ago