The frequency of a wave is 10 Hz. If the speed of the wave is 5 m/s, what is the wavelength? A. 0.5 m B. 2 m C. 10 m D. 50 m

An ice cube at 0c was dropped into 30.0 g of water in a cup at 45.0c. at the instant that all of the ice was melted, the tempera

Ede4ka [16]

The amount of heat given by the water to the block of ice can be calculated by using
$Q=m_w C_{sw} \Delta T_w$
where
$m_w = 30 g$ is the mass of the water
$C_{sw}=4.18 J/(g ^{\circ}C)$ is the specific heat capacity of water
$\Delta T_w = 45.0^{\circ}-19.5^{\circ}C = 20.5^{\circ}C$ is the variation of temperature of the water.

Using these numbers, we find
$Q=(30 g)(4.18 J/(g^{\circ}C))(20.5^{\circ}C)=2571 J$

This is the amount of heat released by the water, but this is exactly equal to the amount of heat absorbed by the ice, used to melt it into water according to the formula:
$Q = m_i L_f$
where $m_i$ is the mass of the ice while $L_f =334 J/g$ is the specific latent heat of fusion of the ice.
Re-arranging this formula and using the heat Q that we found previously, we can calculate the mass of the ice:
$m_i = \frac{Q}{L_f}= \frac{2571 J}{334 J/g} =7.7 g$

3 0

3 years ago

A gymnast with mass m1 = 43 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 =

Liula [17]

A gymnast with mass m1 = 43 kg is on a balance beam that sits on (but is not attached to) two supports. The beam has a mass m2 = 115 kg and length L = 5 m. Each support is 1/3 of the way from each end. Initially the gymnast stands at the left end of the beam.
1)What is the force the left support exerts on the beam?
2)What is the force the right support exerts on the beam?
3)How much extra mass could the gymnast hold before the beam begins to tip?
Now the gymnast (not holding any additional mass) walks directly above the right support.

4)What is the force the left support exerts on the beam?
5)What is the force the right support exerts on the beam?

6 0

3 years ago

A car travels from point A to point B, moving in the same direction but with a non-constant speed. The first half of the distanc

Dmitrij [34]

Answer:

Explanation:

From A to B

distance traveled with velocity $v_1$ in time $t_1$

$\frac{d}{2}=v_1t_1----1$

from B to C

distance traveled is 0.5 d with $v_2$ and $v_3$ velocity for half-half time

$\frac{d}{2}=\frac{v_2t_2}{2}+\frac{v_3t_3}{2}----2$

divide 1 and 2 we get

$\frac{1}{1}=\frac{2v_1t_1}{v_2t_2+v_3t_3}$

$\frac{t_1}{t_2}=\frac{v_2+v_3}{2v_1}$

Now average velocity is given by

$v_{avg}=\frac{d}{t_1+t_2}$

taking $t_1$ common

$v_{avg}=\frac{2v_1t_1}{t_1(1+\frac{t_2}{t_1})}$

$v_{avg}=\frac{2v_1}{1+\frac{2v_1}{v_2+v_3}}$

$v_{avg}=\frac{2v_1(v_2+v_3)}{2v_1+v_2+v_3}$

6 0

3 years ago

The magnitude of each force is 208 N the force on the right is applied at an angle 36° and the mass of the block is 17 kg the co

djyliett [7]

Answer:

11.06m/s²

Explanation:

According to Newtons second law of motion

$\sm F_x = ma_x\\F_m - F_f = ma_x\\mgsin \theta - \mu R mgcos \theta = ma_x\\$

Given

Mass m = 17kg

Fm = 208N

theta = 36 degrees

g = 9.8m/s²

a is the acceleration

Substitute

208 - 0.148(17)(9.8)cos 36 = 17a

208 - 24.6568cos36 = 17a

208 - 19.9478 = 17a

188.05 = 17a

a = 188.05/17

a = 11.06m/s²

Hence the the magnitude of the resulting acceleration is 11.06m/s²

6 0

2 years ago

Three cars (car F, car G, and car H) are moving with the same velocity when the driver suddenly slams on the brakes, locking the

Nastasia [14]

Answer:

Car H

Explanation:

Frictional force is a resistant force. It is given as:

F = u*m*g

Where u = coefficient of friction

m = mass

g = acceleration due to gravity

From the formula above, we see that frictional force is dependent on the mass of object and the coefficient of friction.

Since they all have the same tires, the coefficient of friction between the tire and the floor is the same for each car. Acceleration due to gravity, g, is constant.

The only factor that determines the frictional force of each car is the mass. Hence, the more the mass, the more the frictional force.

So, the most massive car will have the most frictional force and hence, will come to a stop quicker than the others. The least massive car will have the least frictional force and so, will take a longer time to stop.

5 0

2 years ago