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JulijaS [17]
3 years ago
14

A car starting from rest (i.e. initial velocity = 0.0 m/s), moves in the positive X-direction with a constant average accelerati

on = 3.1 m/s2 for 7.9 s. What will be its velocity at that time.
Physics
1 answer:
olga_2 [115]3 years ago
8 0

Answer:

Final speed of the car, v = 24.49 m/s

Explanation:

It is given that,

Initial velocity of the car, u = 0

Acceleration, a=3.1\ m/s^2

Time taken, t = 7.9 s

We need to find the final velocity of the car. Let it is given by v. It can be calculated using first equation of motion as :

v=u+at

v=0+3.1\times 7.9

v = 24.49 m/s

So, the final speed of the car is 24.49 m/s. Hence, this is the required solution.

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A bolt is dropped from a bridge under construction, falling 94 m to the valley below the bridge. (a) how much time does it take
gregori [183]
Refer to the diagram shown below.

When the bolt is dropped at a height of 94 m, its initial velocity, V, is zero.
The last 26% of its fall is at a height of 0.26*94 = 24.4 m.
At that time, the bolt has fallen by 94 - 24.4 = 69.56 m.

The time, t, for the bolt to fall a known distance obeys the equation 
s = Vt + (1/2)gt²,
where
s = 69.56 m, vertical distance traveled, and
g = acceleration due to gravity.

Therefore
69.56 = 0 + (1/2)*9.8*t²
t² = (69.56*2)/9.8
t = 3.7677 s

The total time, T, to fall 94 m is given by 
94 = (1/2)*9.8*T^2
T² = 19.1837
T = 4.38 s

The time taken to pass through the last 26% of its fall is
T - t = 4.38 - 3.7677 = 0.6122 s

The speed after falling 69.56 m is given by
V₁ = 0 + g*t = 36.9235 m/s

The speed with which the bolt strikes the ground is given by
V₂ = 0 +g*T = 9.8*4.38 = 42.924 m/s

Answer:
(a) The bolt takes 0.6122 s to pass through the last 26% of its fall.
(b) When the bolt begins the last 26% of its fall, its speed is 36.92 m/s (nearest hundredth).
(c) Just before it strikes the ground, the speed of the bolt is 42.94 m/s (nearest hundredth).

8 0
3 years ago
A ball is shot at an angle of 45 degrees into the air with initial velocity of 46 ft/sec. Assuming no air resistance, how high d
mestny [16]

Answer:

5.02 m

Explanation:

Applying the formula of maximum height of a projectile,

H = U²sin²Ф/2g...................... Equation 1

Where H = maximum height, U = initial velocity, Ф = angle, g = acceleration due to gravity.

Given: U = 46 ft/sec = 14.021 m/s, Ф = 45°

Constant: g = 9.8 m/s²

Substitute these values into equation 1

H = (14.021)²sin²45/(2×9.8)

H = 196.5884×0.5/19.6

H = 5.02 m.

Hence the ball goes 5.02 m high

8 0
3 years ago
2 forces are acting on a body. I forgot the body doesn't it's position or shape, the forces.......a)must be of equal magnitude b
Veseljchak [2.6K]

Answer:

C

Explanation:

only if there is a net force of zero, the body will not move

some people may say B but that is wrong because maybe one force is greater than the other so the object would still move even though the forces are in opposite directions and parallel

5 0
4 years ago
A bird is flying with through the air with a speed of 18.0 m/s. As it flies, it holds in its claws a 2 kg fish that it
Serggg [28]
It’s started with wing
8 0
3 years ago
Read 2 more answers
A woman is running around a playground while keeping an eye on her child who is on a swing set. The child is going back and fort
Nutka1998 [239]

Answer:

(a) 1500 m

(b) 2827.43m

Explanation:

Given that the time for one cycle of the swing is 1 s

The radius of the swing, R= 3.0m

The angle covered, \theta, by each swing is a quarter of the circle. i.e.

\Theta=\frac {\pi}{2}

Speed of running of women =5 m/s.

Time of running = 5 minutes= 5 x 60 secondes= 300 s.

(a) As distance= (speed) x (time)

So, the required distance= 5 x 300 m= 1500 m.

(b) As there are two swings in one cycle, so, the distance covered in one swing is the length of the circular shown in the figure.

As arc length, l= \theta R, where \theta is the angle, in radian, subtended by the arc at the center, and R is the radius of curvature of the arc.

So, the distance covered by the child in 1 swing = \frac {\pi}{2}\times 3 m=\frac{3\pi}{2}m.

In 1 cycle, there are 2 swings, so distance covered in 1 cycle = 3 \pi m.

Now, in 1 second there is 1 cycle, so in 5 minutes there will be 300 cycles.

So, the total distance covered by the child

= 3\pi\times 300 m= 2827.43 m.

3 0
4 years ago
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