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erastovalidia [21]
4 years ago
9

A small car of mass m and a large car of mass 4m drive along a highway at constant speeds VS and VL. They approach a curve of ra

dius R. The small and large cars have accelerations as and aL respectively, as they travel around the curve. The magnitude of as is twice of that of aL. How does the speed of the small car VS compare to the speed of the large car VL as they move around the curve
Physics
1 answer:
Arte-miy333 [17]4 years ago
4 0

Answer:

v_S=\sqrt{2}v_L

Explanation:

The acceleration experimented while taking a curve is the centripetal acceleration a=\frac{v^2}{r}. Since a_S=2a_L, we have that: \frac{v_S^2}{r_S}=\frac{2v_L^2}{r_L}

They take the same curve, so we have: r_S=r_L=R

Which means: v_S^2=2v_L^2

And finally we obtain: v_S=\sqrt{2}v_L

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A rock falls off a cliff that is 82 m high. How fast is the rock moving when it hits the ground?
drek231 [11]

Answer:

40.1m/s

Explanation:

This is a kinematics problem

x = 82m

a = -9.8m/s^2

v(initial) = 0

Find: v(final)

Using kinematics equation:

vf^2 = vi^2 + 2a(delta)x

vf^2 = 0 + 2*(-9.8)*(82)

vf = sqrt(1607.2) = 40.09m/s

The is traveling with a velocity of 40.1m/s when it hits the ground.

7 0
3 years ago
A circuit consists of a series combination of 5.50 âkΩ and 5.00 âkΩ resistors connected across a 50.0-V battery having negligi
omeli [17]

Answer:

(a) 18.87 V

(b) 23.81 V

(c) 20.75%

Explanation:

The answers are given in the pictures. I have attached the pictures because circuits were needed to be drawn which are easier to understand when done on page. The page are numbered on top left corner.

7 0
4 years ago
An 80-kg quarterback jumps straight up in the air right before throwing a 0.43-kg football horizontally at 15 m/s . How fast wil
viktelen [127]

Answer:

a

The speed of the quarterback backward is v_q =  0.08 \ m/s

b

Known are

 m_Q , m_B , (v_{Bx})_i  (v_{Qx})_f, (v_{Bx})_f

Unknown

   (v_{Qx})_f

Explanation:

From the question we are told that

   The mass of the quarterback is m_Q =  80 \ kg

    The mass of the ball is m_B =  0.43 \ kg

     The speed of the ball is  v_{B x}=  15 \ m/s

The law of momentum conservation can be mathematically represented as

       m_Q u_{Qx} + m_Bu_{Bx}  =  - m_{Q} v_{Qx} + m_B v_{Bx}

Now at initial both ball and quarterback are at rest and the negative sign signify that the quarterback moved backwards after throwing the ball

  So

       m_Q v_{Qx} =  m_B v_ {Bx}

=>     v_{Qx} =  \frac{m_Bv_{Bx}}{m_Q}

substituting values

        v_q =  \frac{0.43 * 15}{80}

       v_q =  0.08 \ m/s

5 0
3 years ago
If planet A is four times as far from the sun as planet C, then the period of its orbit will be__ times as long
Burka [1]
Within the system of the same star, the period of a planet's orbit is
proportional to the 3/2 power of its distance from the central body.
(Kepler's empirical third law of planetary motion, promoted to being
etched in stone by Newton's gravitation.)

(4) ^ 3/2 = <em>8 times</em> as long.
4 0
4 years ago
Long, long ago, on a planet far, far away, a physics experiment was carried out. First, a 0.210-kg ball with zero net charge was
tigry1 [53]

Answer:

\Delta V=316167V

Explanation:

The difference of electric potential between two points is given by the formula \Delta V=Ed, where <em>d</em> is the distance between them and<em> E</em> the electric field in that region, assuming it's constant.

The electric field formula is E=\frac{F}{q}, where <em>F </em>is the force experimented by a charge <em>q </em>placed in it.

Putting this together we have \Delta V=\frac{Fd}{q}, so we need to obtain the electric force the charged ball is experimenting.

On the second drop, the ball takes more time to reach the ground, this means that the electric force is opposite to its weight <em>W</em>, giving a net force N=W-F. On the first drop only <em>W</em> acts, while on the second drop is <em>N</em> that acts.

Using the equation for accelerated motion (departing from rest) d=\frac{at^2}{2}, so we can get the accelerations for each drop (1 and 2) and relate them to the forces by writting:

a_1=\frac{2d}{t_1^2}

a_2=\frac{2d}{t_2^2}

These relate with the forces by Newton's 2nd Law:

W=ma_1

N=ma_2

Putting all together:

N=W-F=ma_1-F=ma_2

Which means:

F=ma_1-ma_2=m(a_1-a_2)=m(\frac{2d}{t_1^2}-\frac{2d}{t_2^2})=2md(\frac{1}{t_1^2}-\frac{1}{t_2^2})

And finally we substitute:

\Delta V=\frac{Fd}{q}=\frac{2md^2}{q}(\frac{1}{t_1^2}-\frac{1}{t_2^2})

Which for our values means:

\Delta V=\frac{2(0.21Kg)(1m)^2}{7.7\times10^{-6}C}(\frac{1}{(0.35s)^2}-\frac{1}{(0.65s)^2})=316167V

7 0
3 years ago
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