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Tresset [83]
3 years ago
10

A car starts from rest and accelerates with a constant acceleration of 1.00 m/s2 for 3.00 s. The car continues for 5.00 s at con

stant velocity. How far has the car traveled from its starting point?
Physics
1 answer:
barxatty [35]3 years ago
8 0

Answer:

19.5 m

Explanation:

Using the equation of motion,

For the first 3 seconds,

s = ut +1/2at².......................... Equation 1

Where t = time, u = initial velocity, a = acceleration, s = distance

Note: Since the car starts from rest, u = 0 m/s.

Given: a = 1.0 m/s, t = 3 s, u = 0 m/s

Substitute into equation 1

s = 0×3+1/2(3)²(1)

s = 9/2

s = 4.5

In the remaining five seconds, at a constant velocity,

s' = vt............. Equation 2

Where v = velocity, t = time.

Recall,

v = u+at

v = 0+1(3)

v = 3 m/s.

Also, t = 5 s.

Substitute into equation 2

s' = 3(5)

s' = 15 m.

Total distance = 15+4.5

Total distance = 19.5 m.

Hence the car travels 19.5 m

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A car starts from rest and increases its speed to 18 m/s in 22 seconds. What is the cars acceleration
IrinaVladis [17]

Magnitude of acceleration = (change of speed) / (time for the change)

Change of speed = (speed at the end) - (speed at the start)

Change of speed = (18 m/s) - (0 m/s) = 18 m/s

Time for the change = 22 sec

Magnitude of acceleration = (18 m/s) / (22 sec)

<em>Magnitude of acceleration = 0.818 m/s²</em>

4 0
4 years ago
A horizontal spring is lying on a frictionless surface. One end of the spring is attaches to a wall while the other end is conne
kipiarov [429]

Answer:

0.832 m/s

Explanation:

The work done by the spring W equals the kinetic energy of the object K

The work done by the spring W = 1/2k(x₀² - x₁²) where k = spring constant, x₀   = initial compression = 0.065 m and x₁ = final compression = 0.032 m

The kinetic energy of the object, K = 1/2mv² where m = mass of object and v = speed of object

Since W = K,

1/2k(x₀² - x₁²) = 1/2mv²

k(x₀² - x₁²) = mv²

mv² = k(x₀² - x₁²)

v² = [(k/m)(x₀² - x₁²)]

taking square root of both sides, we have

v = √[(k/m)(x₀² - x₁²)] since ω = angular frequency = √(k/m),

v = √[(k/m)√(x₀² - x₁²)]

v = ω√(x₀² - x₁²)]

Since ω = 14.7 rad/s, we substitute the other variables into the equation, so we have

v = 14.7 rad/s × √((0.065 m)² - (0.032 m)²)]

v = 14.7 rad/s × √(0.004225 m² - 0.001024 m²)]

v = 14.7 rad/s × √(0.003201 m²)

v = 14.7 rad/s × 0.056577

v = 0.832 m/s

8 0
3 years ago
A 20 N force is applied to an object causing it to move 10 m. How much work was done on the object? How much energy was needed t
olga2289 [7]

Answer:

Here, force=20N and displacement=10m

Work=Force×Displacement=20N×10m=200Nm

3 0
3 years ago
C. Two persons A and B are pushing a block from opposite sides with
Stels [109]

Answer:

1 N

Explanation:

1) You'll just have to sum the force vectors, and since they are opposite, you'll calculate the difference of their magnitudes, in absolute value.

Ftot = |F1 - F2| = 6N - 5N = 1N

2)The direction will be the same as the greater force's direction.

5 0
2 years ago
Please help with these questions as well! I need urgent help! I will give brainliest! God bless!
Radda [10]

6.  Since we are not sure if the person in the question is actively lifting the crate, we have to determine the downwards force of the crate due to gravity and compare it to the normal force.  

F = ma

F = (15.3)(-9.8)

F = -150N

Since the downwards force of the crate is equivalent to the normal force, it means the person is applying no force in picking up the object.  So to pick up a 150N object from scratch, you would have to exert more force than the weight of the object, so the answer is 294N.


7.  Same idea as question 2.  

First determine the weight of the object:

F = ma

F = (30)(-9.8)

F = -294N

The crate in question is not moving, so the magnitudes of the forces in the upwards and downwards direction has to equal to 0.

-294 + 150N + x = 0

x = 144N  

So the person is exerting 144 N.


10.  First find the force of block B to the right due to its acceleration:

F = ma

F = (24)(0.5)

F = 12N

So block B is moving 12N to the right relative to block A due to block A's movement to the left.  However, block A is being applied a much greater force and is moving quicker to the left than block B is moving to the right of bock A.  The force that is causing block B to experience the lower relative force to the right is because of the friction.  To find the friction:

The sum of the forces in the leftward and rightward direction for block B must equal 12N.

75 - x = 12

x = 63N

So the force of friction of block A on block B is 63N to the left.


5 0
3 years ago
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