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2 years ago

Sublimation is used to describe

2 answers:

6 0

Sublimation is the transition of a substance directly from the solid to the gas state, without passing through the liquid state; therefore:

c. the process of ice (solid) changing into water vapor (gas)

hope this helps:)

NNADVOKAT [17]2 years ago

5 0

The answer is c as this phase change refers to any change of a solid to gas.

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A cart traveling at 0.3 m/s collides with stationary object. After the collision, the cart rebounds in the opposite direction. T

padilas [110]

Answer:

Explanation:

First case

A cart speed is 0.3m/s. i.e the initial velocity is u=0.3m/s

It collide with a stationary body, then after collision the ball rebounds and move in opposite direction. This shows that the ball have a velocity after impulse let say v

Then, impulse is given as the change in linear momentum of a body

Impulse =m∆v

I=m(v-u)

Note, momentum is a vector quantity.

I=m(v--u)

I=m(v+u)

I=m(v+0.3)

I¹=0.3m+mv. Equation 1

Second case

A cart speed is 0.3m/s. i.e the initial velocity is u=0.3m/s

It collide with a stationary body, then after collision the ball is at rest, this show that the final velocity is v=0

Then, impulse is given as the change in linear momentum of a body

Impulse =m∆v

I=m(v-u)

Note, momentum is a vector quantity.

I=m(v--u)

I=m(v+u)

In this case v=0 u=0.3m/s

I=m(0+0.3)

I²=0.3m. Equation 2

If we compare impulse 1 (I¹) to impulse 2 (I²)

Subtract equation 2 from 1

We have, I¹ - I² =0.3m+mv -0.3m

I¹ - I² =mv

I¹ =mv+I²

We notice that the first impulse (I¹) is greater than second impulse (I²) by mv.

The correct answer is A

5 0

3 years ago

50 POINTS! A Boy throws a ball horizontally a distance of 22m downrange from the top of a tower that is 20.0m tall. What is his

DerKrebs [107]

The ball's horizontal and vertical velocities at time $t$ are

$v_x=v_{xi}$

$v_y=v_{yi}-gt$

but the ball is thrown horizontally, so $v_{yi}=0$ . Its horizontal and vertical positions at time $t$ are

$x=v_{xi}t$

$y=20.0\,\mathrm m-\dfrac g2t^2$

The ball travels 22 m horizontally from where it was thrown, so

$22\,\mathrm m=v_{xi}t$

from which we find the time it takes for the ball to land on the ground is

$t=\dfrac{22\,\rm m}{v_{xi}}$

When it lands, $y=0$ and

$0=20.0\,\mathrm m-\dfrac{9.8\frac{\rm m}{\mathrm s^2}}2\left(\dfrac{22\,\rm m}{v_{xi}}\right)^2$

$\implies v_i=v_{xi}=11\dfrac{\rm m}{\rm s}$

7 0

3 years ago

A cake is removed from a 375°F oven and placed on a cooling rack in a 63°F room. After 30 minutes the cake is 175°F. When will i

marysya [2.9K]

Answer:

The cake will be at temperature 150°F at after 37.34 minutes

Explanation:

Let T be the temperature of the cake at any time

T∞ be the temperature around the cooling rack = 63°F

T₀ be the initial temperature of the cake = 375°F

And m, c, h are all constants from the cooling law relation

From Newton's law of cooling

Rate of Heat loss by the cake = Rate of Heat gain by the environment

- mc (d/dt)(T - T∞) = h (T - T∞)

(d/dt) (T - T∞) = dT/dt (Because T∞ is a constant)

dT/dt = (-h/mc) (T - T∞)

Let (h/mc) be k

dT/(T - T∞) = -kdt

Integrating the left hand side from T₀ to T and the right hand side from 0 to t

In [(T - T∞)/(T₀ - T∞)] = -kt

(T - T∞)/(T₀ - T∞) = e⁻ᵏᵗ

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

Inserting the known variables

(T - 63) = (375 - 63)e⁻ᵏᵗ

(T - 63) = 312 e⁻ᵏᵗ

At 30 minutes, T = 175°F

175 - 63 = 312 e⁻ᵏᵗ

112/312 = e⁻ᵏᵗ

- kt = In (112/312) = In (0.3590)

- 30k = - 1.025

k = 1.025/30 = 0.0342 /min

When the temp is 150°F,

(T - T∞) = (T₀ - T∞)e⁻ᵏᵗ

(150 - 63) = 312 e⁻ᵏᵗ

e⁻ᵏᵗ = (87/312) = 0.2788

- kt = In 0.2788 = - 1.277

t = 1.277/k = 1.277/0.0342 = 37.34 min.

7 0

4 years ago

A 208-volt, 4-wire, 3-phase wye power system feeds a school building. Daily voltage readings taken for 1 week are 205.25 volts,

joja [24]

For a 208-volt, 4-wire, 3-phase wye power system feeds a school building, the average daily voltage is mathematically given as

x=206.23v

<h3> What is the average daily voltage of the system?</h3>

Generally, the equation for the mean is mathematically given as

x=total sum/ number

Therefore

ts=05.25 volts, 203.75 volts, 208 volts, 204.35 volts, 206.7 volts, 207 volts, and 208.55 volts.

ts=1433.6

In conclusion

x=1433.6/7

x=206.23v

Read more about mean

brainly.com/question/20118982

3 0

3 years ago

A 250-kN railroad car A is traveling at 40 m/s while a 550-kN freight car B is traveling at 10 m/s (both cars heading to the rig

Damm [24]

Answer:

the maximum deformation undergone by the spring = 47.46 cm

Explanation:

Using conservation of momentum:

$m_Av_A + m_Bv_B = (m_A+m_B )v$

where:

$m_A = \frac{250\ kN}{g}$

$v_A = 40$

$m_B = \frac{550\ kN}{g}$

$v_B =10$

Then;

$m_Av_A + m_Bv_B = (m_A+m_B )v$

$\frac{250*10^3}{9.81}*40 + \frac{550*10^3}{9.81}*10 = (\frac{800*10^3}{9.81} )v$

$1580020.387 = 81549.43935 \ v$

$v = \frac{1580020.387}{81549.43935}$

v = 19.375 m/s

However ; using conservation of energy to determine the maximum deformation undergone by the spring ; we have:

$\frac{1}{2} [m_Av_A^2 +m_Bv_B^2] =\frac{1}{2}[(m_A+m_B)v^2 + kx^2]$

$[m_Av_A^2 +m_Bv_B^2] =[(m_A+m_B)v^2 + kx^2]$

$[\frac{250*10^3}{9.81}*40^2 + \frac{550*10^3}{9.81}*10^2] =[ (\frac{800*10^3}{9.81} )*19.375^2 + 70 *10^6 \ * x^2]$

x = 0.4746 m

x = 47.46 cm

Thus, the maximum deformation undergone by the spring = 47.46 cm

3 0

4 years ago

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