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ankoles [38]
2 years ago
14

When starting a car some energy is used to increase the fuels temperature during ignition.this energy causing combustion is a:ab

sorbed energyb:chemical energy c:activation energy d:latent energy
Physics
1 answer:
s2008m [1.1K]2 years ago
5 0
Internal combustion engine works by heat. The chemical energy of the fuel turns into heat energy when the fuel is burned, which produces mechanical energy to drive the pistons, to manage the column of the facility and run the vehicle on the road.
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What does decelerate mean
fenix001 [56]
To slow down or reduce speed.
5 0
3 years ago
A block is held at rest against a wall by a force of magnitude F exerted at an angle theta from the horizontal, as shown in the
wel

Answers:

B.) F cos\theta=F_{n}

C.) F sin\theta=F_{g} \pm F_{f}

Explanation:

The image attached shows the way the force F is acting on the block. Now, if we draw a free body diagram of the situation and write the equations for the Net Force in X and Y, we will have the following:

Net Force in X:

-F_{n}+F cos\theta=0 (1)

Where:

F_{n} is the Normal force

F is the magnitude of the force exerted on the block

\theta is the angle

Net Force in Y:

F sin\theta \pm F_{f}-F_{g}=0 (2)

Where:

F_{f} is the Friction force (it is expresed with the \pm sign because this force may be up or down, we cannot know because the block is at rest)

F_{g} is the gravity force

Rewrittin (1):

F cos\theta=F_{n} (3) This is according to option B

Rewritting (2):

F sin\theta=F_{g}\pm F_{f} (3) This is according to option C

3 0
3 years ago
What is the humidity if the dry-bulb is 10℃ and the wet-bulb is 6℃?
LenaWriter [7]

Answer:

Hello,~There!

What is the humidity if the dry-bulb is 10℃ and the wet-bulb is 6℃?

<h2><u>33% According to the Graph</u></h2>

Hope this helps!

6 0
2 years ago
A horizontal spring with spring constant 210 Nm is compressed by 20 cm and then used to launch a 250 g box across the floor. The
Damm [24]

Answer:

ugmd = 1/2 kx²

d = (1/2 kx²) / (ugm)

= (1/2 * 250 N/m * (0.2 m)²) / (0.23 * 9.81 m/s² * 0.3 kg)

= 7.4 m

ugmd = 1/2 mv²

v = √2ugd

= √(2(0.23)(9.81 m/s²)(7.4 m)

= 5.8 m/s

Explanation:

3 0
3 years ago
A dolphin in an aquatic show jumps straight up out of the water at a velocity of 15.0 m/s. (a) List the knowns in this problem.
astra-53 [7]

Answer:

a)

Y0 = 0 m

Vy0 = 15 m/s

ay = -9.81 m/s^2

b) 7.71 m

c) 3.06 s

Explanation:

The knowns are that the initial vertical speed (at t = 0 s) is 15 m/s upwards. Also at that time the dolphin is coming out of the water, so its initial position is 0 m. And since we can safely assume this happens in Earth, the acceleration is the acceleration of gravity, which is 9.81 m/s^2 pointing downwards

Y(0) = 0 m

Vy(0) = 15 m/s

ay = -9.81 m/s^2 (negative because it points down)

Since acceleration is constant we can use the equation for uniformly accelerated movement:

Y(t) = Y0 + Vy0 * t + 1/2 * a * t^2

To find the highest point we do the first time derivative (this is the speed:

V(t) = Vy0 + a * t

We equate this to zero

0 = Vy0 + a * t

0 = 15 - 9.81 * t

15 = 9.81 * t

t = 0.654 s

At this time it will have a height of:

Y(0.654) = 0 + 15 * 0.654 - 1/2 * 9.81 * 0.654^2 = 7.71 m

The doplhin jumps and falls back into the water, when it falls again it position will be 0 again. So we can equate the position to zero to find how long it was in the air knowing that it started the jump at t = 0s.

0 = Y0 + Vy0 * t + 1/2 * a * t^2

0 = 0 + 15 * t - 1/2 * 9.81 t^2

0 = 15 * t - 4.9 * t^2

0 = t * (15 - 4.9 * t)

t1 = 0 This is the moment it jumped into the air

0 = 15 - 4.9 * t2

15 = 4.9 * t2

t2 = 3.06 s This is the moment when it falls again.

3.06 - 0 = 3.06 s

5 0
2 years ago
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