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2 years ago

Which statement is true regarding the relationship between distance and displacement

2 answers:

7 0

Distance is a scalar and equals to the total distance travelled.
Displacement is a vector and equals to the change in position. Final position - Initial position. The relationship between distance and displacement is that it is motion. The object is in motion and changes his position over time.

SpyIntel [72]2 years ago

5 0

Distance is always either the same or greater than the displacement for the same trip. Distance can never be less than displacement.

You haven't listed any choices so it's not possible to help you choose the correct one.

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Mrs. Rohaley gives you a piece of iron and asks you to determine the physical and chemical properties of the metal. She wants yo

Fofino [41]

Answer:The answer is B

Explanation:

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3 years ago

Concept Simulation 20.4 provides background for this problem and gives you the opportunity to verify your answer graphically. Ho

77julia77 [94]

Answer:

The time constant is 1.049.

Explanation:

Given that,

Charge $q{t}= 0.65 q_{0}$

We need to calculate the time constant

Using expression for charging in a RC circuit

$q(t)=q_{0}[1-e^{-(\dfrac{t}{RC})}]$

Where, $\dfrac{t}{RC}$ = time constant

Put the value into the formula

$0.65q_{0}=q_{0}[1-e^{-(\dfrac{t}{RC})}]$

$1-e^{-(\dfrac{t}{RC})}=0.65$

$e^{-(\dfrac{t}{RC})}=0.35$

$-\dfrac{t}{RC}=ln (0.35)$

$-\dfrac{t}{RC}=-1.049$

$\dfrac{t}{RC}=1.049$

Hence, The time constant is 1.049.

6 0

2 years ago

A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich t

Evgen [1.6K]

Answers:

a) $65.075 kgm/s$

b) $10.526 s$

c) $61.82 N$

Explanation:

<h3>a) Impulse delivered to the ball</h3>

According to the Impulse-Momentum theorem we have the following:

$I=\Delta p=p_{2}-p_{1}$ (1)

Where:

$I$ is the impulse

$\Delta p$ is the change in momentum

$p_{2}=mV_{2}$ is the final momentum of the ball with mass $m=0.685 kg$ and final velocity (to the right) $V_{2}=57 m/s$

$p_{1}=mV_{1}$ is the initial momentum of the ball with initial velocity (to the left) $V_{1}=-38 m/s$

So:

$I=\Delta p=mV_{2}-mV_{1}$ (2)

$I=\Delta p=m(V_{2}-V_{1})$ (3)

$I=\Delta p=0.685 kg (57 m/s-(-38 m/s))$ (4)

$I=\Delta p=65.075 kg m/s$ (5)

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately $1.0 cm=0.01 m$ :

$V_{2}=V_{1}+at$ (6)

$V_{2}^{2}=V_{1}^{2}+2ad$ (7)

Where:

$a$ is the acceleration

$d=0.01 m$ is the length the ball was compressed

$t$ is the time

Finding $a$ from (7):

$a=\frac{V_{2}^{2}-V_{1}^{2}}{2d}$ (8)

$a=\frac{(57 m/s)^{2}-(-38 m/s)^{2}}{2(0.01 m)}$ (9)

$a=90.25 m/s^{2}$ (10)

Substituting (10) in (6):

$57 m/s=-38 m/s+(90.25 m/s^{2})t$ (11)

Finding $t$ :

$t=1.052 s$ (12)

<h3>c) Force applied to the ball by the bat </h3>

According to Newton's second law of motion, the force $F$ is proportional to the variation of momentum $\Delta p$ in time $\Delta t$ :

$F=\frac{\Delta p}{\Delta t}$ (13)

$F=\frac{65.075 kgm/s}{1.052 s}$ (14)

Finally:

$F=61.82 N$

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2 years ago

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3 years ago

The diagram shows the movement of air as a result of convection currents. At which point is the air at its highest density?

Gnesinka [82]

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2 years ago

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