Ask question

Ask question

All categories

2 years ago

10

A block of ice is sliding down a ramp of slope 45° to the horizontal. At the bottom of the ramp, the block strikes a wall with a

force of 3.4 N. What is the mass of the ice? Assume the force of friction is not significant.

1 answer:

laiz [17]2 years ago

8 0

Answer:

Mass, m = 0.49 kg

Explanation:

It is given that,

A block of ice is sliding down a ramp of slope 45° to the horizontal. At the bottom of the ramp, the block strikes a wall with a force of 3.4 N.

We need to find the mass of the ice.

On a sloping surface, the force with which it strikes is given by :

$F=mg\cos\theta\\\\m=\dfrac{F}{g\cos\theta}\\\\m=\dfrac{3.4}{9.8\times \cos(45)}\\\\m=0.49\ kg$

So, the mass of the ice block is 0.49 kg.

You might be interested in

A constant 10.0-N horizontal force is applied to a 20.0-kg cart at rest on a level floor. If friction is negligible, what is the

Otrada [13]

Answer:

2.83 m/s

Explanation:

Given:

Mass of the cart (m) = 20.0 kg

Initial velocity of the cart (u) = 0 m/s

Final velocity (v) = ? m/s

Displacement of the cart (S) = 8.0 m

Horizontal force acting on the cart (F) = 10.0 N

Surface is frictionless. So, only horizontal force is the force acting on the cart.

Now, as per work-energy theorem, the work done by the net force acting on a body is equal to the change in the kinetic energy of the body.

Here, the work done by the horizontal force is given as:

$W_{net}=FS=(10\ N)(8.0\ m)=80\ Nm$

The change in kinetic energy is given as:

$\Delta K=\frac{1}{2}m(v^2-u^2)\\\\\Delta K=\frac{1}{2}(20.0\ kg)(v^2-0)\\\\\Delta K=10v^2$

Now, from work-energy theorem:

$\Delta K=W_{net}\\\\10v^2=80\\\\v^2=\frac{80}{10}\\\\v=\sqrt{8}=2.83\ m/s$

Therefore, the speed of the cart when it has been pushed 8.0 m is 2.83 m/s.

5 0

2 years ago

The acceleration of a particle traveling along a straight line isa=14s1/2m/s2, wheresis in meters. Ifv= 0,s= 1 m whent= 0, deter

Yuliya22 [10]

Answer:

0.78m/s

Explanation:

We are given that

Acceleration= $a=\frac{1}{4}s^{\frac{1}{2}}m/s^2$

v=0, s=1 when t=0

We have to find the particle's velocity at s=2m

We know that

$a=\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=\frac{dv}{ds}v$

$vdv=ads$

$\int_{0}^{v} vdv=\int_{1}^{s}0.25s^{\frac{1}{2}}ds$

$\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2})^{s}_{1}$

By using formula: $\int x^ndx=\frac{x^{n+1}}{n+1}+C$

$\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2}}-1$

Substitute s=2

$\frac{v^2}{2}=\frac{0.50}{3}((2^{1.5})-1)$

$\frac{v^2}{2}=\frac{0.50}{3}\times 1.83$

$v^2=2\times 0.305=0.61$

$v=\sqrt{0.61}=0.78m/s$

Hence, the velocity of particle at s=2m=0.78m/s

3 0

3 years ago

You have a rod with a length of 146.4 cm. You prop up one end on a brick which is 3.8 cm thick. Your uncertainty in measuring th

AfilCa [17]

Answer:

$\partial \theta = 0.003$

Explanation:

we know that

$sin\theta = \frac{3.8}{146.4}$

$\theta = sin^{-1} \frac{3.8}{146.4}$

$\theta = 1.484°$

$\theta = 1.484° *\frac{\pi}{180} = 0.0259 radians$

as we see that $sin\theta = \theta$

relative error $\frac{\partial \theta}{\theta} = \frac{\partial X}{X_1} +\frac{\partial X}{X_2}$

Where X_1 IS HEIGHT OF ROCK

$X_2$ IS THE HEIGHT OF ROAD

$\partial X$ = uncertainity in measuring distance

$\partial X = 0.05$

Putting all value to get uncertainity in angle

$\frac{\partial \theta}{0.0259} = \frac{0.05}{3.8} +\frac{0.05}{146.4}$

solving for $\partial \theta$ we get

$\partial \theta = 0.003$

3 0

2 years ago

What is the energy level of hydrogen?

barxatty [35]

Answer:

The energies corresponding to each of the allowed orbitals are called energy levels.

Explanation:

A scientist known as Niels Bohr put forward that electrons in an atom covers some permitted orbitals with a specific energy. In other words, the energy of an electron in an atom is not continuous, but 'quantized.' The energies corresponding to each of the allowed orbitals are called energy levels.

$E = -\frac{E_0}{n^2} \\where \\E_0 = 13.6 eV (1 eV = 1.602\times 10^{-19}Joules)\\and\ n = 1,2,3...$

8 0

2 years ago

How do you do to your question to see if its answered?

Jobisdone [24]

Answer:

Go in notifications, it'll show if it was answered. If it doesn't show that a person answered it, wait a while, someone might respond :)

Explanation:

On the Top right of your screen, theres a bell button. Click that and it will show all the notifications. it will also show if a person answered it.

It will pop up like

*random username* answered your question! ]

Hope this helped

6 0

2 years ago

Read 2 more answers