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Mandarinka [93]
3 years ago
15

PLEASE ANSWER...... Suppose a car is traveling in the negative x-direction and comes to a stop. What is the sign of that car’s a

cceleration? How do you know?
Physics
2 answers:
Alborosie3 years ago
5 0

positive - accn is slowing car moving in neg x

neg if car in pos x

iren [92.7K]3 years ago
4 0

Answer:

The sign of acceleration is positive

Explanation:

For the car to stop, there must be a deceleration, which will gradually reduce its speed to zero, the deceleration is the aceleration in the opposite way (sign) to the movement, since in this way it will be contrary to the car's displacement and it will be able to stop it.

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Light travelling in one material enters another material in which it travels faster. The light wave will:
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The light will change its direction and bend away from the normal.
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A 0.20-kg particle moves along the x axis under the influence of a conservative force. The potential energy is given by
anastassius [24]

Answer:

e) 11 m/s

Explanation:

For a particle under the action of a conservative force, its mechanical energy at point 0 is must be equal to its mechanical energy at point 1:

K_1+U_1=K_0+U_0\\\\\frac{mv_1^2}{2}+[(8.0\frac{J}{m^2})(x_1)^2+(2.0\frac{J}{m^4})(x_1)^4]=\frac{mv_0^2}{2}+[(8.0\frac{J}{m^2})(x_0)^2+(2.0\frac{J}{m^4})(x_0)^4]

In x_1=1.0m the speed is given, so v_1=5.0\frac{m}{s} and x_0=0. Replacing:

\frac{mv_1^2}{2}+[(8.0\frac{J}{m^2})(1m)^2+(2.0\frac{J}{m^4})(1m)^4]=\frac{mv_0^2}{2}+[(8.0\frac{J}{m^2})0^2+(2.0\frac{J}{m^4})(0)^4]\\\frac{mv_1^2}{2}+8.0J+2.0J=\frac{mv_0^2}{2}\\\frac{mv_0^2}{2}=\frac{mv_1^2}{2}+10J\\v_0=\sqrt{\frac{2}{m}(\frac{mv_1^2}{2}+10J)}\\v_0=\sqrt{\frac{2}{0.2kg}(\frac{(0.2kg)(5\frac{m}{s})^2}{2}+10J)}\\v_0=11.18\frac{m}{s}

7 0
3 years ago
Deer ticks can carry both Lyme disease and human granulocytic ehrlichiosis (HGE). In a study of ticks in the Midwest, it was fou
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Answer:

a)0.024

b)0.148

Explanation:

Let 's represent the set of deer ticks Carrying Lyme disease with L and the set of deer ticks carrying Human Granulocytic Ehrlichiosis with H

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P(H) = 0.10

P(L n H) = 0.1 ·P( L u H )

Hence, P( L u H) = 10 ·P( L nH)

(a)

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Hence, 10 · P(L n H ) = 0.16 + 0.1 - P(L n H )

Hence, 11 · P(L n H) = 0.16 + 0.1 = 0.26

Hence, P(L n H) = 0.26/11=0.024

(b)

We know that condition probability P(H ║ L) = p(L n H)/P(L)

hence, P(H ║ L) =(0.26/11)/0.16 =0.148

3 0
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Answer:

B AND D

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