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zhenek [66]
2 years ago
7

What is the most commmon injury in the United states

Physics
1 answer:
Bess [88]2 years ago
5 0
Home injury’s or at home falls
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Suppose that you are headed toward a plateau 4040 m high. If the angle of elevation to the top of the plateau is 2020degrees°​,
Lelu [443]

Answer :

The distance is 109.89 m.

Explanation :

Given that,

Height = 40 m

Angle = 20°

We need to calculate the base of the​ plateau

Using formula of angle

\tan\thata=\dfrac{h}{x}

Where, h = height

x = base

\theta = angle

Put the value into the formula

\tan20^{\circ}=\dfrac{40}{x}

x=\dfrac{40}{tan20^{\circ}}

x =109.89\ m

Hence, The distance is 109.89 m.

3 0
3 years ago
Behavior that benefits others is called behavior
Ratling [72]
Prosocial Behavior.
5 0
3 years ago
Suppose that a simple pendulum consists of a small 60.0 g bob at the end of a cord of negligible mass. If the angle 0 between th
erik [133]

Based on the mass of the bob and the angle between the cord and the vertical, the pendulum length is 0.50m.

The maximum kinetic energy can be found to be 9.42 x 10⁻⁴J.

<h3>What is the pendulum length?</h3>

This can be found as:

= g-force / w²

Solving gives:

= 9.8 / 4.43²

= 0.4998 m

= 0.50 m

<h3>What is the maximum kinetic energy?</h3>

This can be found as:

= 0.5 × m × w² × A²

Maximum kinetic energy is:

= 0.5 × 60 × 10⁻³ × (4.43 × 0.4998 x 0.08 rad)²

= 9.42 x 10⁻⁴J

Find out more on maximum kinetic energy at brainly.com/question/24690095.

5 0
1 year ago
what is the average speed of (a) a car that travels 400m in 20s. and (b) an athlete who runs 1500m in 4 minutes​
zmey [24]

Answer:

a) 20 m/s

b) 37.5 m)s

Explanation:

Average speed = total distance ÷ total time

=> (a) average speed of a car that travels 400m in 20s

= 400/20 = 20 m/s

& (b) average speed of an athlete who runs 1500m in 4 minutes (or 4×60=240 seconds)

= 1500/240 = 37.5 m/s

5 0
3 years ago
A team of engineering students is designing a catapult to launch a small ball at A so that it lands in the box. If it is known t
Murljashka [212]

Answer:

Explanation:

If the initial velocity is U

Then the horizontal component of the velocity is

Ux= Ucosθ

Then the range for a projectile is give as

R=Ux.t

Where t is the time of flight

The time of flight is given as

t=2USinθ/g

Therefore,

R=Ux.t

R=UCosθ.2USinθ/g

R=U^2×2SinθCosθ/g

Then, from trigonometric ratio

2SinθCosθ= Sin2θ

R=U^2Sin2θ/g

Given that θ=32° and g=9.81m/s^2

Then

R=U^2Sin2×32/9.81

R=U^2Sin64/9.81

R=0.0916U^2

Then, range is given by R=0.0916U^2

A=0.0916U^2.

T

The box is at a distance A from the point of projection. Then the range R=A

R=0.0916U^2

A=0.0916U^2

Then,

U^2=A/0.0916

U^2=10.915A

Then the initial velocity should be

U=√10.915A

U=3.3√A

8 0
3 years ago
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