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svp [43]
3 years ago
13

Ms. Stafford wants to try racing with a push start. A student pushes her at 5 m/s before the rocket kicks in. The rocket still o

nly accelerates her at a rate of 4 m/s/s. Calculate how many meters she travels in 7 seconds after the rocket kicks in.
Physics
1 answer:
Gnom [1K]3 years ago
8 0
The initial velocity of Ms. Stafford is v_0 = 5 m/s, while her acceleration is 
a=4 m/s^2
This is a uniform accelerated motion, so we can calculate the total distance travelled by Ms. Stafford in a time of t=7.0 s using the law of motion for a uniform accelerated motion:
S=v_0t +  \frac{1}{2} at^2 = (5 m/s)(7.0 s)+ \frac{1}{2}(2 m/s^2)(7.0 s)^2 = 84 m
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NNADVOKAT [17]
When you are finding work, the easiest way is to use the formula.

W = F*D

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4 0
3 years ago
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You took a running leap off a high diving platform. You were running at 3.1 m/s and hit the water 2.4 seconds later. How high wa
rusak2 [61]

The distance you free-fall from rest is  D = (1/2) (g) (T²) <== memorize this

Height of the platform = (1/2) (9.8 m/s²) (2.4 sec)²

Height = (4.9 m/s²) (5.76 s²)

Height = (4.9/5.76) meters

Height = 28.2 meters (a VERY high platform ... about 93 ft off the water !)

Without air-resistance, your horizontal speed doesn't change.  It's constant.  Traveling 3.1 m/s for 2.4 sec, you cover (3.1 m/s x 2/4 s) = 7.4 m horizontally.

7 0
3 years ago
A real battery with internal resistance 0.460 Ω and emf 9.00 V is used to charge a 56.0-µF capacitor. A 21.0-Ω resistor is put i
Margaret [11]

Answer: 1.176×10^-3 s

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t =RC

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t = 56×10^-6 × 21

t = 1176×10^-6

t = 1.176×10^-3 s

4 0
3 years ago
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Friction occurs when the ____ and ____ of two surfaces grind against each other.
monitta

Friction occurs when the surfaces and heat of two surfaces grind against each other.

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7 0
3 years ago
A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are
Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of x= 5t^3+1

Coordinate of position of y=3t^2+2

Time = 2.00 s

We need to calculate the acceleration

a = \dfrac{d^2x}{dt^2}

For x coordinates

x=5t^3+1

On differentiate w.r.to t

\dfrac{dx}{dt}=15t^2+0

On differentiate again w.r.to t

\dfrac{d^2x}{dt^2}=30t

The acceleration in x axis at 2 sec

a = 60i

For y coordinates

y=3t^2+2

On differentiate w.r.to t

\dfrac{dy}{dt}=6t+0

On differentiate again w.r.to t

\dfrac{d^2y}{dt^2}=6

The acceleration in y axis at 2 sec

a = 6j

The acceleration is

a=60i+6j

We need to calculate the net force

F = ma

F = 3.00\times(60i+6j)

F=180i+18j

The magnitude of the force

|F|=\sqrt{(180)^2+(18)^2}

|F|=180.89\ N

Hence, The net force acting on this object is 180.89 N.

3 0
3 years ago
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