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borishaifa [10]
2 years ago
13

What is Otter's average velocity over his entire trip when it takes him 2 minutes to walk 100 meters north and another 1 minute

to walk 70 meters south?Express your answer using the proper SI units. Round your answer to two decimal places, and include a direction if necessary.
Physics
1 answer:
Leya [2.2K]2 years ago
7 0

Answer:

0.50m/s

Explanation:

Average velocity is the change in displacement of a body with respect to time.

Velocity = ∆S/∆t

∆S = 100m - 70m

∆S = 30m

∆t = 2min - 1 min

∆t = 1min = 60secs

Substitute the given parameters into the formula for velocity

Velocity = 30m/60s

Velocity = 1/2 m/s

Average Velocity = 0.5m/s

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A cyclist going downhill is accelerating at 1.2 m/s2. If the final velocity of the cyclist is 16 m/s after 10 seconds, what is t
Blababa [14]

Answer:

Initial Velocity is 4 m/s

Explanation:

What is acceleration?

It is the change in velocity with respect to time, or the rate of change of velocity.

We can write this as:

a=\frac{\Delta v}{t}

Where

a is the acceleration

v is velocity

t is time

\Delta  is "change in"

For this problem , we are given

a = 1.2

t = 10

Putting into formula, we get:

a=\frac{\Delta v}{t}\\1.2=\frac{\Delta v}{10}\\\Delta v = 1.2*10\\\Delta v = 12

So, the change in velocity is 12 m/s

The change in velocity can also be written as:

\Delta v = Final  \ Velocity - Initial \ Velocity

It is given Final Velocity = 16, so we put it into formula and find Initial Velocity. Shown Below:

\Delta v = Final  \ Velocity - Initial \ Velocity\\12=16-Initial \ Velocity\\Initial \ Velocity = 16 - 12 = 4

hence,

Initial Velocity is 4 m/s

3 0
3 years ago
Read 2 more answers
A 25N force is applied to a bar that can pivot around its end. The force is r=0.75 m away from the end of an angle at 0= 30. wha
Alecsey [184]

Answer:

<h2>9.375Nm</h2>

Explanation:

The formula for calculating torque τ = Frsin∅ where;

F = applied force (in newton)

r = radius (in metres)

∅ = angle that the force made with the bar.

Given  F= 25N, r = 0.75m and ∅ = 30°

torque on the bar τ  = 25*0.75*sin30°

τ = 25*0.75*0.5

τ = 9.375Nm

The torque on the bar is 9.375Nm

6 0
2 years ago
A 2.0-kg object is lifted vertically through 3.00 m by a 150-N force. How much work is done on the object by gravity during this
noname [10]

Answer:

-58.8 J

Explanation:

The work done by a force is given by:

W=Fdcos \theta

where

F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the direction of the force and the displacement.

In this problem, we are asked to find the work done by gravity, so we must calculate the magnitude of the force of gravity first, which is equal to the weight of the object:

F=mg=(2.0 kg)(9.8 m/s^2)=19.6 N

The displacement of the object is d = 3.00 m, while \theta=180^{\circ}, because the displacement is upward, while the force of gravity is downward; therefore, the work done by gravity is

W=Fdcos \theta=(19.6 N)(3.00 m)(cos 180^{\circ})=-58.8 J

And the work done is negative, because it is done against the motion of the object.


6 0
2 years ago
Read 2 more answers
A visitor to the observation deck of a skyscraper manages to drop a penny over the edge. As the penny falls faster, the force du
pentagon [3]
If a coin is dropped at a relatively low altitude, it's acceleration remains constant. However, if the coin is dropped at a very high altitude, air resistance will have a significant effect. The initial acceleration of the coin will be the greatest. As it falls down, air resistance will counteract the weight of the coin. So, the acceleration will decrease. Although the acceleration decreases, the coin still accelerates, that is why it falls faster. When the air resistance fully counters the weight of the coin, the acceleration will become zero and the coin will fall at a constant speed (terminal velocity). So, the answer should be, The acceleration decreases until it reaches 0. The closest answer is.
a. The acceleration decreases.
8 0
2 years ago
Read 2 more answers
A sample of an unknown volatile liquid was injected into a Dumas flask (mflask = 27.0928 g, Vflask = 0.1040 L) and heated until
NNADVOKAT [17]

Answer:

The gas was Hexane

Explanation:

taking the diference between the mass of the flask and the final mass qe can calculate the mass of liquid injected (assuming none escaped the flask):

m_{l}  = 27.4593g - 27.0928g = 0.3665g

with the volume of the flask we can get the density of the gas at the indicated pressure and temperature:

d_{g}  = \frac{0.3665 g}{0.1040L} = 3.524 g/L

From the ideal gases law we have that the density can be calculated as:

d_{g}  = \frac{P*M}{R*T}

Where R is the ideal gases constant = , and M the molecular weight of the fluid. Solving for M:

M=\frac{d_{g}*R*T}{P}=\frac{3.524g/L*0,082atmL/molK*291K}{0.976atm}

M=86.16 g/mol

Note that the temperature is computed in Kelvin T= 18+273=291K

The gas with the closer molar mass is Hexane

4 0
2 years ago
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