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Pepsi [2]
3 years ago
7

An athlete competing in long jump leaves the ground with a speed of 9.14 m/s at an angle of 35.00 above the horizontal. What is

the length of the athlete's jump?
Physics
1 answer:
Lemur [1.5K]3 years ago
5 0

Answer:

      R = 8.01 m

Explanation:

We can solve this problem using the projectile launch equations. The jump length is the throw range

           R = v₀² sin  2θ / g

in the exercise they give us the initial speed of 9.14 m / s and in the launch angle 35º

let's calculate

           R = 9.14² sin (2  35) / 9.8

           R = 8.01 m

this is the jump length

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How much water can be held by a cylindrical tank with a radius of 12 feet and a height of 30 feet? A. 2,260.8 cubic feet B. 13,5
MakcuM [25]

Answer:

Volume, V = 13564.8  cubic feet

Explanation:

It is given that,

Radius of the cylindrical tank, r = 12 feet

Height of the tank, h = 30 feet

We need to find the water that can be held by a cylindrical tank i.e. we need to find the volume of the tank. It is given by :

V=\pi\times r^2\times h

V=3.14\times (12)^2\times 30

V = 13564.8  cubic feet

So, the water held by the tank is 13564.8  cubic feet. Hence, this is the required solution.

6 0
3 years ago
Air pollutants often cause irritation in the _____ <br> system.
erastova [34]
Respiratory system.

Oversimplified Explanation: they enter the lungs, which is part of the respiratory system.
7 0
3 years ago
A spherical balloon has a radius of 6.55 m and is filled with helium. The density of helium is 0.179 kg/m3, and the density of a
Ksju [112]

Answer:

M_1 = 317.7 kg

Explanation:

Mass of the helium gas filled inside the volume of balloon is given as

m = \rho V

m = 0.179(\frac{4}{3}\pi R^3)

m = 0.179(\frac{4}{3}\pi 6.55^3)

m = 210.7 kg

now total mass of balloon + helium inside balloon is given as

M = 210.7 + 990

M = 1200.7 kg

now we know that total weight of balloon + cargo = buoyancy force on the balloon

so we will have

(M + M_1)g = \rho_{air} V g

(1200.7 + M_1) = (\frac{4}{3}\pi 6.55^3) (1.29)

1200.7 + M_1 = 1518.4

M_1 = 317.7 kg

3 0
4 years ago
A 60 N force is applied over a distance of 15 m. How much work was done?
GuDViN [60]

Answer:

900J

Explanation:

w =f×s

60×15

=900J

thus the k.e of the body is 900j

3 0
3 years ago
Consider a situation where a constant force of 25 N acts on an object having a mass of 2 kg for 3 seconds. What is the work done
blagie [28]

Answer:

Work done W =1406.25 J

Explanation:

Work done on a body can be calculated using newton's 2nd laws:

F=ma

\Rightarrow a=\frac{F}{m}

Hence acceleration of the block is given by:

\Rightarrow a=\frac{25}{2}=12.5m/s^2

Displacement of the object is given by:

\Rightarrow S=ut+\frac{1}{2}at^2

Substitute the values

\Rightarrow S=0*3+\frac{1}{2}(12.5)3^2

\Rightarrow S=56.25 m

Now work done is given by:

 W=F.S

W = 25×56.25

W =1406.25 J

3 0
3 years ago
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