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artcher [175]
4 years ago
7

Tom kicks a soccer ball on a flat, level field giving it an initial speed of 20 m/s at an angle of 35 degrees above the horizont

al. a) How long will the ball be in the air? b) What maximum height will the ball attain? c) How far away from Tom will the ball land? d) What speed will the ball have in the instant just before it lands?
Physics
1 answer:
SVEN [57.7K]4 years ago
8 0

Answer:

(a) 2.34 s

(b) 6.71 m

(c) 38.35 m

(d) 20 m/s

Explanation:

u = 20 m/s, theta = 35 degree

(a) The formula for the time of flight is given by

T = \frac{2 u Sin\theta }{g}

T = \frac{2 \times 20 \times Sin35 }{9.8}

T = 2.34 second

(b) The formula for the maximum height is given by

H = \frac{u^{2} \times Sin^{2}\theta }{2g}

H = \frac{20^{2} \times Sin^{2}35 }{2 \times 9.8}

H  = 6.71 m

(c) The formula for the range is given by

R = \frac{u^{2} \times Sin 2\theta }{g}

R = \frac{20^{2} \times Sin 2 \times 35}{9.8}

R = 38.35 m

(d) It hits with the same speed at the initial speed.

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umka21 [38]

height = .0970 mm = .00831 inches

Volume = length * breadth * height = 8.5 * 11 * 0.00381

Volume = 0.776985 inches^cube

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3 years ago
If a company like Waste Management collects all of the paper, bottles, plastic, ete in a single bin, how
Tju [1.3M]

Answer:

yes they can be recycled

Explanation:

8 0
4 years ago
An airplane accelerates from a speed of 88m/s to a speed of 132 m/s during a 15 second time interval. How far did the airplane t
Gelneren [198K]

Answer:

1650\:\mathrm{m}

Explanation:

We can use the following kinematics equations to solve this problem:

v_f=v_i+at,\\{v_f}^2={v_i}^2+2a\Delta x.

Using the first one to solve for acceleration:

132=88+a(15),\\15a=44,\\a=\frac{44}{15}=2.9\bar{3}\:\mathrm{m/s^2}.

Now we can use the second equation to solve for the distance travelled by the airplane:

132^2=88^2+2\cdot2.9\bar{3}\cdot \Delta x,\\\Delta x= \frac{9680}{2\cdot2.9\bar{3}},\\\Delta x =\fbox{$ 1650\:\mathrm{m}$}(three significant figures).

6 0
3 years ago
A donkey is attached by a rope to a wooden cart at an angle of 23° to the horizontal. the tension in the rope is 210 n. if the c
kaheart [24]

Answer:

787528.7 J

Explanation:

Work done: This can be defined as the product of force and distance along the direction of force. The S.I unit of work is Joules (J).

From the question,

W = Tcos∅(d)............. Equation 1

Where W = work done, T = tension in the rope, ∅ = the angle of the rope to the horizontal, d = distance.

But,

d = v(t)..................... equation 2

Where v = velocity, t = time

Substitute equation 2 into equation 1

W = Tcos∅(vt)............. Equation 3

Given: T = 210 N, ∅ = 23°, v = 7 km/h = 1.94 m/s, t = 35 min = 2100 s

Substitute into equation 3

W = 210(cos23°)(1.94×2100)

W = 787528.7 J

8 0
3 years ago
A weightlifter lifts a set of weights a vertical distance of 2 meters. If a constant net force of 350 N is exerted on the weight
o-na [289]

Answer:

The work done by the weightlifter, W = 700 J

The power of the weightlifter, P = 350 watts

Explanation:

A weightlifter lifts a set of weights a vertical distance, s = 2 m

The force exerted to lift the weight, F = 350 N

The work done by the body is defined as the product of the force applied by the body to the displacement it caused.

                            W = F x s

                                = 350 N x 2 m

                                = 700 J

The work done by the weightlifter, W = 700 J

The time taken by the weightlifter to lift the weight, t = 2 s

The power is defined as the rate of body to do work. It is given by the equation,

                            P = W / t

                               = 700 J / 2 s

                               = 350 watts

Hence, the power of the weightlifter, P = 350 watts

6 0
3 years ago
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