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Korvikt [17]
3 years ago
10

Two cars, C and D, travel in the same direction on a long, straight section of highway. During a particular time interval Ato, c

ar D is ahead of car C and is speeding up while car C is slowing down. During the interval At, it is observed that car C gains on car D (i.e., the distance between the cars decreases). Explain how this is possible, and give a specific example of such a case.
Physics
1 answer:
a_sh-v [17]3 years ago
6 0

Answer:

possibly because the car is running out of gas

Explanation:

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How much work is done by the force lifting a 0.1-kilogram hamburger vertically upward at a constant velocity 0.3 meter from a ta
pychu [463]

Answer:

0.2943 Nm

Explanation:

Work done is given a the product of force and diatance moved and expressed by the formula

W=Fd

Here W represent work, F is applied force and d is perpendicular distance

Also, we know that F=mg where m is the mass of an object and g is acceleration due to gravity. Substituting this back into the initial equation then

W=mgd

Taking acceleration due to gravity as 9.81 m/s2 and substituting mass with 0.1 kg and distance with 0.3 m then

W=0.1*9.81*0.3=0.2943 Nm

5 0
3 years ago
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Schach [20]

Answer:

c

Explanation:

7 0
2 years ago
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A projectile is launched horizontally from a 20-m tall edifice with a vox of 25 m/s. How long will it take for the projectile to
NISA [10]

Answer:

a) First let's analyze the vertical problem:

When the projectile is on the air, the only vertical force acting on it is the gravitational force, then the acceleration of the projectile is the gravitational acceleration, and we can write this as:

a(t) = -9.8m/s^2

To get the vertical velocity we need to integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

where v0 is the initial vertical velocity because the object is thrown horizontally, we do not have any initial vertical velocity, then v0 = 0m/s

v(t) = (-9.8m/s^2)*t

To get the vertical position equation we need to integrate over time again, to get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + p0

where p0 is the initial position, in this case is the height of the edifice, 20m

then:

p(t) = (-4.9m/s^2)*t^2+ 20m

The projectile will hit the ground when p(t) = 0m, then we need to solve:

(-4.9m/s^2)*t^2+ 20m = 0m

20m = (4.9m/s^2)*t^2

√(20m/ (4.9m/s^2)) = t = 2.02 seconds

The correct option is a.

b) The range will be the total horizontal distance traveled by the projectile, as we do not have any horizontal force, we know that the horizontal velocity is 25 m/s constant.

Now we can use the relationship:

distance = speed*time

We know that the projectile travels for 2.02 seconds, then the total distance that it travels is:

distance = 2.02s*25m/s = 50.5m

Here the correct option is a.

c) Again, the horizontal velocity never changes, is 25m/s constantly, then here the correct option is option b. 25m/s

d) Here we need to evaluate the velocity equation in t = 2.02 seconds, this is the velocity of the projectile when it hits the ground.

v(2.02s) =  (-9.8m/s^2)*2.02s = -19.796 m/s

The velocity is negative because it goes down, and it matches with option d, so I suppose that the correct option here is option d (because the sign depends on how you think the problem)

4 0
3 years ago
As the bright sun shines upon the water, the water slowly disappears. The same sunlight gives energy to the surrounding plans to
mariarad [96]

first is physical .... sec chem

8 0
3 years ago
Confirm if this is correct or not. If it isn't correct, please correct it.
kow [346]

Answer:

d = 421.83 m

Explanation:

It is given that,

Height, h = 396.9 m

Horizontal speed, v = 46.87 m/s

We need to find the distance traveled by the ball horizontally. Let t is the time taken by the ball. Using second equation of motion for vertical direction. So,

396.9=0\times t+\dfrac{1}{2}\times 9.8 t^2\\\\t=9\ s

Now d is the distance covered by the cannonball. So,

d=vt\\\\d=46.87\times 9\\\\d=421.83\ m

Hence, this is the required solution.

3 0
3 years ago
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