1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Rashid [163]
1 year ago
10

Pls help me with a physics question I am struggling

Physics
1 answer:
konstantin123 [22]1 year ago
7 0

Answer:

It will condese to 47

Explanation:

As per as the table chart , Substance G melting point at 47 and boiling point at 120 so AS per as the chart

You might be interested in
A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7
maxonik [38]
First, let's find the speed v_i of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
p_i = m_1 v_1
After the collision, the two blocks stick together and so now they have mass m_1 +m_2 and they are moving with speed v_i:
p_f = (m_1 + m_2)v_i
For conservation of momentum
p_i=p_f
So we can write
m_1 v_1 = (m_1 +m_2)v_i
From which we find
v_i =  \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force \mu (m_1+m_2)g. The work done by the frictional force to stop the two blocks is
\mu (m_1+m_2)g  d
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
\frac{1}{2} (m_1+m_2)v_i^2
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g  d
From which we can find the value of the coefficient of kinetic friction:
\mu =  \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49
3 0
3 years ago
The charges Q1=Q and Q2=4Q that are a distance d apart, repel each other with a force of 1.60 N. What would be the force between
gladu [14]

Answer:

50.4 N

Explanation:

Q1 = Q

Q2 = 4 Q

Distance = d

The force is given by

F = \frac{KQ_{1}Q_{2}}{d^{2}}

1.60 = \frac{4KQ^{2}}{d^{2}}    .... (1)

Now,

Q3 = 2 Q

Q4 = 7 Q

distance = d/3

F' = \frac{9KQ_{3}Q_{4}}{d^{2}}

F' = \frac{126KQ^{2}}{d^{2}}   .... (2)

Divide equation (2) by equation (1), we get

F' / 1.60 = 126 / 4

F' = 50.4 N

Thus, the force is 50.4 N.

7 0
3 years ago
Calculate the most probable speed of an ozone molecule in the stratosphere
Marysya12 [62]

Answer:

v_{mp}=305.83 m/s

Explanation:

The temperature in stratosphere is generally about 270 K

molecular weight of an ozone molecule = 48 gm/mole

now formula for most probable velocity

v_{mp}= \sqrt{\frac{2RT}{M} }

plugging the values we get

v_{mp}= \sqrt{\frac{2\8.314\times270}{48} }

v_{mp}=305.83 m/s

7 0
3 years ago
A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
atroni [7]

(a) 2.79 rev/s^2

The angular acceleration can be calculated by using the following equation:

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha is the angular acceleration

\theta=50.0 rev is the number of revolutions made by the disk while accelerating

Solving the equation for \alpha, we find

\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

\theta=? is the number of revolutions made by the disk while accelerating

Solving the equation for \theta, we find

\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev

4 0
3 years ago
Which graph accurately shows the relationship between kinetic energy and speed as speed increases?
mixer [17]

Answer:

B

Explanation:

kinetic energy (KE) is the energy possessed by moving bodies. It can be expressed as:

KE = \frac{1}{2}mv^{2}

Where: m is the mass of the object, and v its speed.

For example, a stone of mass 2kg was thrown and moves with a speed of 3 m/s. Determine the kinetic energy of the stone.

Thus,

KE =  \frac{1}{2} x 2 x 3^{2}

     = 9

KE = 9.0 Joules

Assume that the speed of the stone was 4 m/s, then its KE would be:

KE =  \frac{1}{2} x 2 x 4^{2}

     = 16

KE = 16.0 Joules

Therefore, it can be observed that as speed increases, the kinetic energy increases. Thus option B is appropriate.

3 0
3 years ago
Other questions:
  • A diver shines light up to the surface of a flat glass-bottomed boat at an angle of 30° relative to the normal. If the index of
    9·1 answer
  • Newton's law of universal gravitation can be applied to
    5·2 answers
  • If We Start With 48 Atoms Of A Radioactive Substance, How Many Would Remain After One Half-life?
    15·1 answer
  • Given that the velocity of blood pumping through the aorta is about 30 cm/s, what is the total current of the blood passing thro
    8·1 answer
  • Suppose 2.10 C of positive charge is distributed evenly throughout a sphere of 1.30-cm radius. 1) What is the charge per unit vo
    15·1 answer
  • B. A car is moving 4.0 m/s to the right. The car begins to accelerate at a rate of 1.5 m/s/s, to the right. After
    10·1 answer
  • What is transmitted by EM waves?
    11·1 answer
  • Billy drops a ball from a height of 1 m. The ball bounces back to a height of 0.8 m, then
    12·1 answer
  • Two blocks of masses 1.0 kg and 2.0 kg, respectively, are pushed by a constant applied force F across a horizontal frictionless
    8·1 answer
  • If a child pulls a sled through the snow with a force of 50 N exerted at an angle of 38° above
    12·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!