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Anna71 [15]
2 years ago
6

One fine summer day a group of students was jumping from a railroad bridge into the river below. They stepped off the bridge wit

h an acceleration (due to gravity) of 9.8 m/s2 and they hit the water 1.5 s later. How high (in meters) was the bridge? Only enter the number, not the units.
Physics
1 answer:
Mrrafil [7]2 years ago
3 0

Answer:

11 m

Explanation:

The following data were obtained from the question:

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) = 1.5 s

Height (h) =..?

The height of the bridge can be obtained by using the following formula:

h =½gt²

h = ½ × 9.8 × 1.5²

= 4.9 × 2.25

= 11.025 ≈ 11 m

Thus, the height of the bridge is approximately 11 m

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Answer:

6v

Explanation:

V=IR

V= 2* 3

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2 years ago
Which statement best define energy
Ksju [112]

Answer:

"The capacity of a system to perform work of any type."

Explanation:

The best statement to describe Energy is:

"The capacity of a system to perform work of any type."

6 0
3 years ago
A metal wire breaks when its tension reaches 100 newton. If the radius and length of the wire were both doubled then it would br
serg [7]

Answer:

200 N

Explanation:

Since Young's modulus for the metal, E = σ/ε where σ = stress = F/A where F = force on metal and A = cross-sectional area, and ε = strain = e/L where e = extension of metal = change in length and L = length of metal wire.

So,  E = σ/ε = FL/eA

Now, since at break extension = e.

So making e subject of the formula, we have

e = FL/EA = FL/Eπr² where r = radius of metal wire

Now, when the radius and length are doubled, we have our extension as e' = F'L'/Eπr'² where F' = new force on metal wire, L' = new length = 2L and r' = new radius = 2r

So, e' = F'(2L)/Eπ(2r)²

e' = 2F'L/4Eπr²

e' = F'L/2Eπr²

Since at breakage, both extensions are the same, e = e'

So,  FL/Eπr² = F'L/2Eπr²

F = F'/2

F' = 2F

Since F = 100 N,

F' = 2 × 100 N = 200 N

So, If the radius and length of the wire were both doubled then it would break when the tension reached 200 Newtons.

7 0
3 years ago
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zmey [24]
The transfer is perpendicular

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3 0
3 years ago
Read 2 more answers
A 0.50-kg object moves in a horizontal circular track with a radius of 2.5 m. An external forceof 3.0 N, always tangent to the t
kodGreya [7K]

The work done by the force is 47.1 J

Explanation:

The work done by a force in moving an object is given by

W=Fd cos \theta (1)

where

F is the magnitude of the force

d is the distance covered by the object

\theta is the angle between the direction of the force and the motion of the object

In this problem, the force applied to the object is

F = 3.0 N

This force is always tangential to the track: this means that at every instant, the force is parallel to the motion of the object, so

\theta=0

And the distance covered is equal to the circumference of the circle, which is:

d=2\pi r=2\pi (2.5 m)=15.7 m

where r = 2.5 m is the radius.

Now we can substitute into eq.(1) to find the work done:

W=(3.0)(15.7)(cos 0)=47.1 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0
3 years ago
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