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Kay [80]
3 years ago
5

Why is the position vs. time graph for an object in free fall a curve?

Physics
1 answer:
Ymorist [56]3 years ago
7 0
Under free fall, the object is falling with a velocity that is increasing uniformly. Since the gradient of position-time graph reflects the velocity, the gradient is increasing, i.e. curve.
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A dentist uses a concave mirror (focal length 2.5 cm) to examine some teeth. If the distance from the object to the mirror is 1.
AysviL [449]

Answer:

2.28

Explanation:

From mirror formula,

1/f = 1/u+1/v .......... Equation 1

Where f = focal length of the mirror, v = image distance, u = object distance.

Note: The focal length mirror is positive.

make v the subject of the equation,

v = fu/(u-f)............ Equation 2

Given: f = 2.5 cm, u = 1.4 cm

Substitute into equation 2

v = 2.5(1.4)/(1.4-2.5)

v = 3.5/-1.1

v = -3.2 cm.

Note: v is negative because it is a virtual image.

But,

Magnification = image distance/object distance

M = v/u

Where M = magnification.

Given: v = 3.2 cm, u = 1.4 cm

M = 3.2/1.4

M = 2.28.

Thus the magnification of the tooth = 2.28.

3 0
4 years ago
A gold wire that is 1.8 mm in diameter and 15 cm long carries a current of 260 mA. How many electrons per second pass a given cr
Musya8 [376]

Answer:

162500000.  

Explanation:

Given that

Diameter of the wire , d= 1.8 mm

The length of the wire ,L = 15 cm

Current ,I = 260 m A

The charge on the electron ,e= 1.6 x 10⁻¹⁹ C

We know that Current I is given as

I=\dfrac{q}{t}

I=Current

q=Charge

t=time

q= I t

q= 260 m t

The total number of electron = n

q= n e

n=\dfrac{260\times 10^{-3}\ t}{1.6\times 10^{-9}}

n=162500000 t

\dfrac{n}{t}=16250000

The number of electron passe per second will be 162500000.

4 0
3 years ago
This is Yolanda's desk. What could Yolanda do to increase the amount of force needed to change the motion of the desk?
Stella [2.4K]

Yolanda might put more items to the desk to make it heavier, requiring more force.

We need to learn more about the force acting on an object in order to locate the solution.

<h3>How can the force that is required to modify the motion be increased?</h3>
  • We are aware that the word for force is,

                            F=ma

where m denotes the object's mass and an its acceleration

  • There are two ways to increase the force required to alter the motion of the table.
  • One is to increase the mass, and the other is to accelerate it more quickly.
  • Otherwise, there will be a lot of friction between the surfaces, making it difficult to move without exerting a lot of force.

We can infer from this that Yolanda could add items to the desk to increase its mass, necessitating the use of additional force.

Learn more about the force here:

brainly.com/question/4075805

#SPJ1

8 0
2 years ago
Show that Ns is the same as kgmls.<br>​
Minchanka [31]

1 N = 1kgm/s²

then,

1Ns = 1kgm/s² * s

1Ns = 1kgm/s

Hence shown.

3 0
3 years ago
A Thomson's gazelle can run at very high speeds, but its acceleration is relatively modest. A reasonable model for the sprint of
frosja888 [35]

1. 27.3 m/s

The velocity of the gazelle at any time is given by:

v=u+at

where

u is the initial velocity

a is the acceleration

t is the time

Here we have:

u = 0 (the gazelle starts from rest)

a=4.2 m/s^2

t = 6.5 s

Substituting the data, we find the gazelle's top speed:

v=0+(4.2)(6.5)=27.3 m/s

2. 3.8 s

The distance covered by the gazelle is

d = 30 m

We know that the gazelle accelerates during the first part of the motion and then it continues at constant speed. We need to find first if the gazelle completes the race during the first part of its motion (accelerated motion); to do this, we can calculate what would be the distance covered by the gazelle before reaching the top speed, after t = 6.5 s:

d'=\frac{1}{2}at^2 = \frac{1}{2}(4.2)(6.5)^2=88.7 m

Which is larger than 30 m: this means that the gazelle covers the 30 m during its accelerated motion. Therefore, we can use again the equation:

d=\frac{1}{2}at^2

And substituting d = 30 m, we find the time:

t=\sqrt{\frac{2d}{a}}=\sqrt{\frac{2(30)}{4.2}}=3.8 s

3. 10.6 s

In this case, the  distance the gazelle must cover is 200 m.

We know that in the first 6.5 s, the gazelle covers a distance of 88.7 m.

In the second part of the motion, the gazelle continues at its top speed, which is:

v = 27.3 m/s

The gazelle still have to cover a distance of

d' = 200-88.7 =111.3 m

Therefore, the time taken to cover this distance is

t'=\frac{d'}{v}=\frac{111.3}{27.3}=4.1 s

So, the total time the gazelle needs to cover 200 m is

t = 6.5 + 4.1 = 10.6 s

6 0
4 years ago
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