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natali 33 [55]
2 years ago
11

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

ture of the water in the cup was 19.5c. what was the mass of ice cube?
Physics
1 answer:
Ede4ka [16]2 years ago
3 0
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
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If the work function of a material is such that red light of wavelength 700 nm just barely initiates the photoelectric effect, w
marishachu [46]

Answer:

2.13 x 10^-19 J or 0.53 eV

Explanation:

cut off wavelength, λo = 700 nm = 700 x 10^-9 m

λ = 400 nm = 400 x 10^-9 m

Use the energy equation

E = \frac{h c}{\lambda _{o}}+K

Where, K be the work function

\frac{h c}{\lambda} = \frac{h c}{\lambda _{o}}+K

K =hc\left ( \frac{1}{\lambda } -\frac{1}{\lambda _{0}}\right )

K =6.63\times 10^{-34}\times 3\times 10^{8}\left ( \frac{1}{4\times 10^{-7} } -\frac{1}{7\times 10^{-7}}\right )

K = 2.13 x 10^-19 J

K = 0.53 eV

3 0
2 years ago
A soccer ball is kicked At 8m/s. It lands on the ground after being in the air for .85 seconds. At what angle is it kicked?
SpyIntel [72]

If ball remains in air for total time T = 0.85 s

this is also known as time of flight

In order to find the time of flight we can use kinematics

\Delta Y = v_y*t + \frac{1}{2}at^2

so for complete motion its displacement in y direction will be zero

0 = v_y* 0.85 + \frac{1}{2}(-9.8)(0.85^2)

0 = v_y*0.85 - 3.54

v_y = 4.165 m/s

now we know that net velocity of the ball is 8 m/s

while is y direction component we got is vy = 4.165 m/s

now by component method we can say

v_y = v sin\theta

4.165 = 8 sin\theta

\theta = sin^{-1}\frac{4.165}{8}

\theta = 31.4^0

so it is projected at an angle of 31.4 degree above horizontal

8 0
3 years ago
you are riding an elevator upwards at 2.20m/s. you have a balance accurately calibrated in newtons. when you hang a mass of 15.0
Nady [450]

Answer:

<h2> 33N</h2>

Explanation:

Note: the unit for acceleration  in m/s^2

Also, the unit of the mass in the problem is assumed to be in kg

We are going to use newtons second law of motion which expresses the weight in an elevator as

F=ma

Given data

Acceleration of the elevator a= 2.2m/s^2

mass m = 15kg

substituting the data in the expression for force is

F=2.2*15

F= 33N

The reading of the balance is 33N

5 0
2 years ago
A wave created by a certain source travels from medium 1 into another medium 2. It is noticed that its velocity is faster in med
poizon [28]
After thorough researching, if a wave created by a certain source travels from medium 1 into another medium 2. It is noticed that its velocity is faster in medium 2 than in medium 1. The student that discussed rightly what happens to the wave as it moves into the medium is the student 1.
3 0
3 years ago
A uniform 1.4-kg rod that is 0.75 m long is suspended at rest from the ceiling by two springs, one at each end of the rod. Both
svetlana [45]

Answer:

7 deg

Explanation:

m = mass of the rod = 1.4 kg

W = weight of the rod = mg = (1.4) (9.8) = 13.72 N

k_{L} = spring constant for left spring = 59 Nm^{-1}

k_{R} = spring constant for right spring = 33 Nm^{-1}

x_{L} = stretch in the left spring

x_{R} = stretch in the right spring

L = length of the rod = 0.75 m

\theta = Angle the rod makes with the horizontal

Using equilibrium of force in vertical direction for left spring

k_{L} x_{L} = (0.5) W\\(59) x_{L} = (0.5) (13.72)\\x_{L} = 0.116 m

Using equilibrium of force in vertical direction for right spring

k_{R} x_{R} = (0.5) W\\(33) x_{R} = (0.5) (13.72)\\x_{R} = 0.208 m

Angle made with the horizontal is given as

\theta = tan^{-1}(\frac{(x_{R} - x_{L})}{L} )\\\theta = tan^{-1}(\frac{(0.208 - 0.116)}{0.75} )\\\theta = 7 deg

3 0
3 years ago
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