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kirill115 [55]
3 years ago
15

An 5kg object is released from rest near the surface of a planet. The vertical position of the object as a function of time is s

hown in the graph. All frictional forces are considered to be negligible. What is the closest approximation of the weight of the object.
a) 300N
b) 30N
c) 5N
d) 150N

Physics
1 answer:
Tom [10]3 years ago
5 0

Answer:

The correct option is b: 30 N.

Explanation:

First, we need to find the acceleration due to gravity (a):

y_{f} - y_{0} = v_{o}t - \frac{1}{2}a(\Delta t)^{2}   (1)

Where:

y_{f}: is the final vertical position (obtained from the graph)

y_{0}: is the initial vertical position (obtained from the graph)

v₀: is the initial speed = 0 (it is released from rest)

Δt: is the variation of time (from the graph)

From the graph, we can take the following values of height and time:

t₀ = 0 s → t_{f} = 5 s

y₀ = 300 m → y_{f} = 225 m

Now, by entering the above values into equation (1) and solving for "a" we have:

a = 2\frac{y_{0} - y_{f}}{(t_{f} - t_{0})^{2}} = 2\frac{300 m - 225 m}{(5 s - 0)^{2}} = 6 m/s^{2}

Finally, the weight of the object is:

W = ma = 5 kg*6 m/s^{2} = 30 N

Therefore, the correct option is b: 30 N.

I hope it helps you!                                                                                                            

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steposvetlana [31]

Answer:

<h3>The answer is 500 km </h3>

Explanation:

The distance covered by an object given it's velocity and time taken can be found by using the formula

<h3>distance = average velocity × time</h3>

From the question

average speed = 250 km/h

time = 2 hrs

We have

distance = 250 × 2

We have the final answer as

<h3>500 km</h3>

Hope this helps you

5 0
2 years ago
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What is the only planet where a day lasts longer than a year?
Ksivusya [100]

Answer:

Venus

Explanation:

Facts about Venus:

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3 0
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What will be the pressure exerterd by the 0bject if 4000n is acting on an area of 50msqure
Ivahew [28]
<h3>Given, </h3>

Force,F = 4000 N

Area,a = 50 m²

<h3>We know that, </h3>

Pressure = Force/Area

★ Putting the values in the above formula,we get:

\sf \rightarrow \: pressure =  \dfrac{4000}{50}

\sf \rightarrow pressure = 80 \: N {m}^{ - 2}

7 0
3 years ago
Acar accelerates from 4 meters/second to 16 meters/second in 4 seconds. The car's acceleration is
s2008m [1.1K]

To understand this question, you need to understand the concept of acceleration first. Have you ever been in a car and noticed that it was getting faster and faster? That "speeding up" of the car is known as acceleration! Acceleration is essentially the rate at which you speed up.

Okay, so we now know what acceleration is. What are its units? The unit of acceleration is the change in velocity over a period of time: \frac{∆v}{t}

If you haven't learned about velocity yet, just think about it as speed for now. The funny-looking triangle, ∆, is a symbol for "the change of." For example, if I started walking at 3 \frac{feet}{second} then sped up to 5 \frac{feet}{second}, then the change in my speed would be 2 \frac{feet}{second}, because I started walking 2 \frac{feet}{second} faster!

Okay, enough with all the explanations. Hopefully, you understand the units now. Let's take a look at the question. A car accelerates from 4 \frac{meters}{second} to 16 \frac{meters}{second}  in 4 seconds. What would the acceleration be? Let's set up an equation:

a = \frac{∆v}{t}

a is the acceleration, ∆v is the change in velocity, and t is the time elapsed.

Now, let's plug in our values! ∆v is the change in velocity, and to find that we simply have to subtract 16 \frac{meters}{second} by 4 \frac{meters}{second}. That makes sense, right? Back to the equation.

a = \frac{∆v}{t}
a = \frac{16-4}{4}

(16 - 4 is the change in velocity, and 4 is the number of seconds the car was accelerating)

a = \frac{12}{4}

a = 3 (\frac{meters}{second^{2}})

We have our answer! The car's acceleration is 3 meters per second^{2}.

(You might be thinking: Wait. Meters per second squared? The reason for that is because acceleration is the rate at which the speed increases! That makes the unit \frac{\frac{meters}{second}}{second}, which can be simplified down to \frac{meters}{second^{2} })

Let me know if you need clarification on anything I explained here!
- breezyツ

6 0
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A series LR circuit contains an emf source of 19 V having no internal resistance, a resistor, a 22 H inductor having no apprecia
masha68 [24]

Answer: R = 394.36ohm

Explanation: In a LR circuit, voltage for a resistor in function of time is given by:

V(t) = \epsilon. e^{-t.\frac{L}{R} }

ε is emf

L is indutance of inductor

R is resistance of resistor

After 4s, emf = 0.8*19, so:

0.8*19 = 19. e^{-4.\frac{22}{R} }

0.8 = e^{-\frac{88}{R} }

ln(0.8) = ln(e^{-\frac{88}{R} })

ln(0.8) = -\frac{88}{R}

R = -\frac{88}{ln(0.8)}

R = 394.36

In this LR circuit, the resistance of the resistor is 394.36ohms.

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