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1 year ago

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

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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:

Explanation:

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

KE = $\frac{1}{2}$ m $v^{2}$

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