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3 years ago

Which of the following is NOT a way that machines provide a mechanical advantage?

2 answers:

4 0

The following choice that is NOT a way that machines provides a mechanical advantage is to change direction. The correct answer is A.

Mumz [18]3 years ago

4 0

i just took the test, A is incorrect.

im guessing its C, but not certain

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Two thermally insulated cylinders, A and B, of equal volume, both equipped with pistons, are connected by a valve. Initially A h

SVEN [57.7K]

Answer:

The final temperature is $T_1 =(0.5)^{\frac{2}{3} }$

The initial and the final temperature are the same

Explanation:

The explanation is shown in the first and second uploaded image

5 0

2 years ago

Someone is trying to balance a (110cm) plank with certain forces.

Mashutka [201]

Answer:

a. i. 30 Nm ii. This moment is a clockwise positive moment.

b. i. 15 Nm ii, This moment is a counter-clockwise negative moment.

c. i. The plank will not balance. ii. The plank would tip up.

d. 150 N

Explanation:

a) Calculate the moment of the 60N force (about O), then name its type.

i. Calculate the moment of the 60N force (about O)

Since moment = Force × perpendicular distance from point of moment ,

M = Fd

Since F = 60 N and d = 50 cm = 0.5 m

M = 60 N × 0.5 m = 30 Nm

ii. Then name its type.

This moment is a clockwise positive moment.

b) Calculate the moment of the 30N force (about O), then name its type.

i. Calculate the moment of the 30N force (about O),

Since moment = Force × perpendicular distance from point of moment ,

M' = F'd'

Since F' = 30 N and d' = 50 cm = 0.5 m

M' = 30 N × 0.5 m = 15 Nm

ii. Then name its type.

This moment is a counter-clockwise negative moment.

c) Will the plank balance? If not, which way will it tip?

i. Will the plank balance?

The plank will balance if the net moment on it is zero

So net moment, M' = positive moment - negative moment = M - M' = 30 Nm - 15 Nm = 15 Nm

Since the net moment on the plank is M" = 15 Nm ≠ 0,<u>the plank will not balance.</u>

ii Which way will it tip?

The plank would tip in the direction of the greater moment since the net moment is positive. <u>This moment tilts the plank in a clockwise direction, so the plank would tip up.</u>

d) What extra force would be needed at (B) to balance the plank?

The extra force must balance the net moment,

So M" = F"d" where F" = force and d" = distance of force from O = 10 cm = 0.10 m

F" = M"/d"

= 15 Nm/0.10 m

= 150 N

4 0

2 years ago

Using the strap at an angle of 31.0° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15.0

lora16 [44]

Answer:

Explanation:

The question is incomplete. Here is the complete question.

Using the strap at an angle of 31° above the horizontal, a Grade 12 Physics student, tired from studying, is dragging his 15 kg school bag across the floor at a constant velocity. (a) If the force of tension in the strap is 51 N, what is the normal force.

Check the diagram related to the question in the attachment below for better understanding.

The normal force is the reaction acting perpendicular to the force of tension in the strap and opposite the weight of the bag. They are the forces acting along the vertical.

The normal force N will be the sum of the force of tension acting along the vertical (Ty) and the weight of the bag (W).

Ty = 15sin31°

Ty = 7.73N

W = mass * acceleration due to gravity

W = 15.0*9.8

W = 147N

The normal force is therefore expressed as;

N = Ty + W

N = 7.73 + 147

N = 154.73N

3 0

3 years ago

A capacitor is connected to an ac generator that has a frequency of 2.3 kHz and produces a rms voltage of 1.5 V. The rms current

miv72 [106K]

Answer:

Explanation:

Given that,

AC frequency of 2.3KHz

f=2.3×10³Hz

Vrms produce is

Vrms=1.5V

Current rms

Irms= 31mA

The capacitor is reconnected to a generator of frequency

f=4.8KHz =4800Hz

The current rms becomes

Irms= 85mA

Vrms=?

Solution

First genrator

The capacitive reactance is given as

Xc=Vrms/Irms

Xc=1.5/31×10^-3

Xc=48.39 ohms

Now, to know the capacitance of the capacitor

Xc=1/2πfC

Then,

C=1/2πfXc

So,

C=1/2×π×2300×48.39

C=1.43×10^-6C

C=1.43μF

Note: the capacitance of the capacitor did not change,

Now for generator two.

The reactance are given as

Xc=1/2πfC

Xc=1/2×π×4800×1.43×10^-6

Xc=23.19ohms

Then,

Vrms2=Irms2 ×Xc

Vrms2=85×10^-3×23.19ohms

Vrms2=1.97V

Vrms2=1.97Volts

7 0

3 years ago

Determine the gravitational field 300km above the surface of the earth. How does this compare to the field on the earth's surfac

Serjik [45]

The strength of the gravitational field is given by:
$g= \frac{GM}{r^2}$
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.

In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
$r=R+h=6370 km+300 km=6670 km= 6.67 \cdot 10^{6} m$

And now we can use the previous equation to calculate the field strength at that altitude:
$g= \frac{GM}{r^2}= \frac{(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2})(5.97 \cdot 10^{24} kg)}{(6.67 \cdot 10^6 m)^2} = 8.95 m/s^2$

And we can see this value is a bit less than the gravitational strength at the surface, which is $g_s = 9.81 m/s^2$ .

4 0

3 years ago