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inn [45]
3 years ago
14

The table shows the changes in velocity for several different cars. Each change takes place over a 10-second interval. Which car

s are either speeding up or slowing down?
Physics
2 answers:
sesenic [268]3 years ago
6 0
Car B, Car C, And Car E Are either speeding up or slowing down
erica [24]3 years ago
3 0

Answer:

B, C, E are the right answers

Explanation:

I got this question on a test and got these as the right answers

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Why do tides happen every 50 minuets?
Ivan
Because of the moon and earth. The earth spins on its axis (it's tilted) and the moon orbits the earth every 28 days.
3 0
4 years ago
PLEASEHELP THANK YOU SM
garri49 [273]

Answer:

I belive it would be the first answer choice.

Explanation:

When the Mass is lower, the ball moves faster so it would have more kintetic energy

7 0
3 years ago
Read 2 more answers
A force of 6 N acts on a 33 kg object for 9 seconds. What is the objects change in velocity?
GuDViN [60]
Acceleration = change in velocity/time
By F = ma,
6 = 33 x change in velocity / 9
change in velocity = +1.636 m/s
4 0
3 years ago
Two hypothetical planets of masses m1 and m2 and radii r1 and r 2 , respectively, are nearly at rest when they are an infinite d
Leto [7]

Answer:

(a) v_1 = m_2\sqrt{\frac{2G}{d(m_1+m_2)} }

v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)} }

(b) Kinetic Energy of planet with mass m₁, is KE₁ =  1.068×10³² J

Kinetic Energy of planet with mass m₂, KE₂ =  2.6696×10³¹ J

Explanation:

Here we have when their distance is d apart

F_{1} = F_{2} =G\frac{m_{1}m_{2}}{d^{2}}

Energy is given by

Energy \,of \,attraction = -G\frac{m_{1}m_{2}}{d}}+\frac{1}{2} m_{1} v^2_1+ \frac{1}{2} m_{2} v^2_2

Conservation of linear momentum gives

m₁·v₁ = m₂·v₂

From which

v₂ =  m₁·v₁/m₂

At equilibrium, we have;

G\frac{m_{1}m_{2}}{d}} = \frac{1}{2} m_{1} v^2_1+ \frac{1}{2} m_{2} v^2_2       which gives

2G{m_{1}m_{2}}= d m_{1} v^2_1+  dm_{2} (\frac{m_1}{m_2}v_1)^2= dv^2_1(m_1+(\frac{m_1}{m_2} )^2)

multiplying both sides by m₂/m₁, we have

2Gm^2_{2}}= dv^2_1 m_2+dm_1v^2_1 =dv^2_1( m_2+m_1)

Such that v₁ = \sqrt{\frac{2Gm^2_2}{d(m_1+m_2)} }

v_1 = m_2\sqrt{\frac{2G}{d(m_1+m_2)} }

Similarly, with v₁ =  m₂·v₂/m₁, we have

G\frac{m_{1}m_{2}}{d}} = \frac{1}{2} m_{1} v^2_1+ \frac{1}{2} m_{2} v^2_2\Rightarrow  2G{m_{1}m_{2}}= dm_{1} (\frac{m_2}{m_1}v_1)^2 +d m_{2} v^2_2= dv^2_2(m_2+(\frac{m_2}{m_1} )^2)

From which we have;

2G{m^2_{1}}= dm_{2} v_2^2 +d m_{1} v^2_2 and

v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)} }

The relative velocity = v₁ + v₂ =v_1+v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)} } + m_2\sqrt{\frac{2G}{d(m_1+m_2)} } = (m_1+m_2)\sqrt{\frac{2G}{d(m_1+m_2)} }

v₁ + v₂ = (m_1+m_2)\sqrt{\frac{2G}{d(m_1+m_2)} }

(b) The kinetic energy KE = \frac{1}{2}mv^2

KE_1= \frac{1}{2} m_{1} v^2_1 \, \, \, KE_2= \frac{1}{2} m_{2} v^2_2

Just before they collide, d = r₁ + r₂ = 3×10⁶+5×10⁶ = 8×10⁶ m

v_1 = 8\times10^{24}\sqrt{\frac{2\times6.67408 \times 10^{-11}} {8\times10^6(2.00\times10^{24}+8.00\times10^{24})} } = 10333.696 m/s

v_2 = 2\times10^{24}\sqrt{\frac{2\times6.67408 \times 10^{-11}} {8\times10^6(2.00\times10^{24}+8.00\times10^{24})} } =2583.424 m/s

KE₁ = 0.5×2.0×10²⁴× 10333.696² =  1.068×10³² J

KE₂ = 0.5×8.0×10²⁴× 2583.424² =  2.6696×10³¹ J.

7 0
3 years ago
The physical property that determines that how easily heat and electricity pass through a material is?
Varvara68 [4.7K]

Answer: Hello! I'm Jungkook. Here is your answer.....

A. conductivity

Explanation:

Hope this helps! Anneyong/Bye!

xoxoKookie

6 0
2 years ago
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