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2 years ago

6

Heat always transfers from substances with _ thermal energy to substances with ___ thermal until both substances reach t

he ____ temperature. Fill in the blanks.

1 answer:

Reika [66]2 years ago

4 0

Heat always transfers from substances with high thermal energy to substances with low thermal until both substances reach the maximum temperature

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A small rubber ball is launched by a compressed-air cannon from ground level with an initial speed of 11.8 m/s directly upward.

Fantom [35]

Answer:

7.09683 m

1.20285 s

2.4057 s

11.8 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² (negative up, positive down)

From equation of motion we have

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-11.8^2}{2\times -9.81}\\\Rightarrow s=7.09683\ m$

The maximum height above the ground that the ball reaches is 7.09683 m

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-11.8}{-9.81}\\\Rightarrow t=1.20285\ s$

Time taken to go up is 1.20285 s it will take the same time to come down so total time taken to reach the ground after it is shot is 1.20285+1.20285 = 2.4057 s

$v=u+at\\\Rightarrow v=0+9.81\times 1.20285\\\Rightarrow v=11.8\ m/s$

The velocity just before it hits the ground is 11.8 m/s

6 0

2 years ago

Three point charges are arranged along the x axis. Charge q1=-4.00nC is located at x= .250 m and q2= 2.40 nC is at the x= -.300m

Umnica [9.8K]

Answer:

q₃=5.3nC

Explanation:

First, we have to calculate the force exerted by the charges q₁ and q₂. To do this, we use the Coulomb's Law:

$F= k\frac{|q_aq_b|}{r^{2} } \\\\\\F_{13}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(-4.00*10^{-9}C)q_3|}{(.250m)^{2} } =576q_3N/C\\\\F_{23}=(9*10^{9} Nm^{2} /C^{2} )\frac{|(2.40*10^{-9}C)q_3|}{(.300m)^{2} } =240q_3N/C\\$

Since we know the net force, we can use this to calculate q₃. As q₁ is at the right side of q₃ and q₁ and q₃ have opposite signs, the force F₁₃ points to the right. In a similar way, as q₂ is at the left side of q₃, and q₂ and q₃ have equal signs, the force F₂₃ points to the right. That means that the resultant net force is the sum of these two forces:

$F_{Net}=F_{13}+F_{23}\\\\4.40*10^{-9} N=576q_3N/C+240q_3N/C\\\\4.40*10^{-6} N=816q_3N/C\\\\\implies q_3=5.3*10^{-9}C=5.3nC$

In words, the value of q₃ must be 5.3nC.

7 0

3 years ago

A person pushes two boxes with a horizontal force F of magnitude of 100 N.

Monica [59]

The magnitude of the action–reaction pair between the two boxes (A and B) will be "18.2 N".

According to the question,

Mass of box A,

$m_A = 9\ kg$

Mass of box B,

$m_B = 2 \ kg$

Horizontal force,

$F_{app} = 100 \ N$

From the Newton's law,

→ $F_{app} = (\frac{F_{app}}{m_A+m_B} )a$

or,

→ $a = \frac{F_{app}}{(m_A+m_B)}$

Bu substituting the values, we get

→ $= \frac{100}{9+2}$

→ $= \frac{100}{11}$

→ $9.10 \ m/s^2$

We can see that between the two boxes, the action-reaction pair exist.

then,

→ $F_{action-reaction} = m_b \ a$

→ $=2\times 9.10$

→ $= 18.2 \ N$ (magnitude)

Thus the above solution is appropriate.

Learn more about the magnitude here:

brainly.com/question/13545862

7 0

2 years ago

Hi can you please help me, im really stuck

Lady bird [3.3K]

I have the answer for A. Since there is blockage in the ear canal, some sound waves may not be able to get through or travel as quickly so you would have trouble hearing

3 0

3 years ago

The altitude of a hang glider is increasing at a rate of 6.75 m/s. At the same time, the shadow of the glider moves along the gr

Rufina [12.5K]

Answer:

16.45 m/s

Explanation:

Let y be the vertical distance and x be the horizontal distance

We are given that

The altitude of hang glider increasing at the rate= $v_y=\frac{dy}{dt}==6.75m/s$

The shadow of the glider moves along the ground at speed= $v_x=\frac{dx}{dt}=15m/s$

We have to find the magnitude of glider's velocity.

We know that

Magnitude of velocity= $v=\sqrt{v^2_x+v^2_y}$

Substitute the values

$v=\sqrt{(15)^2+(6.75)^2}$

$v=\sqrt{225+45.5625}$

$v=16.45m/s$

Hence, the magnitude of glider's velocity=16.45 m/s

4 0

3 years ago