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Elenna [48]
1 year ago
9

In the diagram below, point A is southwest of point B. Jaleh walks for 400 seconds along path 1, going from point A to point B.

She then runs for 100 seconds from point B back to point A, following path 3. What is Jaleh's velocity for the first part of her trip? (Remember that velocity is calculated using displacement and time.)
A. 0 m/s

B. 0.5 m/s northeast

C. 1.4 m/s southwest

D. 1 m/s northeast

Physics
2 answers:
Maslowich1 year ago
6 0
The answer is indeed A according to my calculations
timama [110]1 year ago
4 0

Answer:

C

Explanation:

velocity = displacement (m) / change of time (s)

velocity = (400 + 300) / (100 + 400)

velocity = 1.4 m/s

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Which of the following choices defines energy in scientific terms?
snow_tiger [21]
1) B. Energy is the ability to do work

2) C. Energy is conserved, it just goes from one form to another.

3) Work = Force x displacement
= 300 x 100 = 30,000 Joules

4) leaning a brick because no displacement is taking place.

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Two flywheels of negligible mass and different radii are bonded together and rotate about a common axis (see below). The smaller
jekas [21]

Answer:

F_2 = 29.54 N

Explanation:

As we know that the combination is maintained at rest position

So we will take net torque on the system to be ZERO

so we know that

\tau = \vec r \times \vec F

here we will have

\vec r_1 \times F_1 = \vec r_2 \times F_2

so we have

13 \times 50 = 22 \times F_2

so we have

F_2 = \frac{13 \times 50}{22}

F_2 = 29.54 N

8 0
3 years ago
An area experiences thunderstorms with high winds and a drop in temperature. Which weather event most likely occurred? A. A stat
gavmur [86]

Answer:

C a fast-moving cold front moved through the area.

Explanation:

This is because, since there is a there is a thunderstorm and high winds in the area, this can only be caused by a fast moving front. Also there is a temperature drop, this can only be caused by the fast moving cold front since a cold front has a low temperature.

Thus, for the area to experience thunderstorms with high winds and a drop in temperature, <u>a fast-moving cold front moved through the area.</u>

5 0
2 years ago
A horizontal 953 N merry-go-round of radius 1.68 m is started from rest by a constant horizontal force of 73.9 N applied tangent
solniwko [45]

Answer:

K.E=365.2 J

Explanation:

Given data

Weight w =953 N

radius r=1.68 m

F=73.9 N

t=2.55 s

g=9.8 m/s²

To find

Kinetic Energy K.E

Solution

From the moment of inertia

I=(1/2)MR^{2}\\ as \\W=mg\\So\\I=(1/2)(W/g)R^{2}\\I=(1/2)(953/9.8)(1.68)^{2}\\I=137.232kg.m^{2}

The angular acceleration is given as

a=T/I\\a=\frac{FR}{I}\\ a=\frac{(73.9)(1.68)}{137.232}\\a=0.905rad/s^{2}

The angular velocity is given as

w=at\\w=(0.905)(2.55)\\w=2.31rad/s

So the Kinetic Energy is given as

K.E=(1/2)Iw^{2}\\ K.E=(1/2)(137.232)(2.31)^{2}\\ K.E=365.2J

3 0
3 years ago
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