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tresset_1 [31]
3 years ago
8

On a vehicle with a manual transmission, the point at which the engine begins to move the car as the clutch is released is calle

d the _____________________
a.matching point

b.blending point

c.friction point

d.friction cuff
Physics
1 answer:
Artyom0805 [142]3 years ago
8 0
The answer to this question is the letter "C" which is the "friction point". Well, to define friction point, it is a part of the vehicle that allows the clutch to connect into the transmission of the vehicle part and initiates the movement of the vehicle. At this point, the engine of the vehicle will start to move.
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According to the barometer in the attachment below, which statement is true?
mixer [17]

Your answer is A

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2 years ago
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_____ are pictures of relationships
butalik [34]

The blank in the question can be filled with the word, “Graph”. Therefore, Graphs are the pictures which are in relationships.

<u>Explanation: </u>

Graph usually represents a set of data which is nonlinear in occurrence and has some relationship between the two given data. And as graph are pictorial representation, it is simply assumed as the pictures of relationships.

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4 0
3 years ago
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A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.
AveGali [126]

Answer:

The magnitude of electric force is  7.2\times10^{-3} N

Explanation:

Coulomb's Law:

The force of attraction or repletion is

  • directly proportional to the products of charges i.e F\propto q_1q_2
  • inversely proportional to the square of distance i.e F\propto \frac{1}{r^2}

\therefore F\propto \frac{q_1q_2}{r^2}

\Rightarrow F=k \frac{q_1q_2}{r^2}    [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force  be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5

we know,

cos \theta = \frac{base }{hypotenuse}

\Rightarrow cos \theta = \frac{4 }{r}

Total force F_Q = 2.F_1 cos\theta \hat{i}

                         =2k\frac{Qq_1}{r^2} cos\theta \hat i

                         =2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i

                         =8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i     [ r=5]

                         =7.2\times10^{-3}\hat i   N

The magnitude of electric force is  7.2\times10^{-3} N

                         

3 0
3 years ago
How does the energy of the marble remain the same at all points on the ramp
Elenna [48]

Answer:

It depends on the potential energy it has ontop of the ramp. The marble has the same potential energy was the kinetic energy without changing the ramp incline or moving it. What kinetic energy shows is what potential energy made it to be, so we look at how the ramp is placed. If the ramp is a steep incline the kinetic energy will be fast. It depends on the weight of the marble too. If the marble is heavy the potential energy will slowly tansition to kinetic energy while a  light marble will transition to kinetic energy fast. But marbles are light so there we have it. It basically goes from potential energy to kinetic energy to thermal energy to potential energy.  Make me brainliest pls :)

Explanation:

7 0
3 years ago
A frog jumps vertically upward from a 20m tall building with an initial velocity of 8.1m/s. How high above the ground will the f
topjm [15]

Answer:

Explanation:

Consider the initial position of the frog (20 m above ground) as the reference position. All measurements are positive measured upward.

Therefore,

u = 10 m/s, initial upward velocity.

H = - 20 m, position of the ground.

g = 9.8 m/s², acceleration due to gravity.

Part (a)

When the frog reaches a maximum height of h from the reference position, its velocity is zero. Therefore

u² - 2gh = 0

h = u²/(2g) = 10²/(2*9.8) = 5.102 m

At maximum height, the frog will be 20 + 5.102 = 25.102 m above ground.

Answer: 25.1 m above ground

Part (b)

Let v = the velocity when the frog hits the ground. Then

v² = u² - 2gH

v² = 10² - 2*9.8*(-20) = 492

v = 22.18 m/s

Answer: The frog hits the ground with a velocity of 22.2 m/s

8 0
3 years ago
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