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3 years ago

Can someone help pleaseeee

Physics

1 answer:

Eduardwww [97]3 years ago

4 0

Answer:

Motion with constant velocity of magnitude 1 m/s (uniform motion) for 4 seconds in a positive direction and then for 2 seconds uniform motion with constant velocity of magnitude 3 m/s in reverse direction .

Explanation:

The graph shows a constant velocity of 1 m/s for 4 seconds in the positive direction. After that, between 4 seconds and 6 seconds, the object reverses its motion with constant velocity of magnitude 3m/s.

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PLEASE HELP ME I DONT UNDERSTAND THIS AND THIS IS TIMED

mixer [17]

(C)

Explanation:

The circle has a radius r = 0.5 m, which means that its circumference C is

$C = 2\pi r = 2\pi(0.5\:\text{m}) = 3.14\:\text{m}$

One revolution means that the stopper travels a distance equal to the circumference of the circle so the velocity of the stopper is

$v = \dfrac{C}{t} =\dfrac{3.14\:\text{m}}{0.2\:\text{s}} = 15.7\:\text{m/s} \approx 16\:\text{m/s}$

5 0

2 years ago

A simple pendulum is used to determine the acceleration due to gravity at the surface of a planet. The pendulum has a length of

SVEN [57.7K]

Answer:

Acceleration due to gravity is 20 $m/sec^2$

So option (E) will be correct answer

Explanation:

We have given length of the pendulum l = 2 m

Time period of the pendulum T = 2 sec

We have to find acceleration due to gravity g

We know that time period of pendulum is given by

$T=2\pi \sqrt{\frac{l}{g}}$

$2=2\times 3.14 \sqrt{\frac{2}{g}}$

$0.3184= \sqrt{\frac{2}{g}}$

Squaring both side

$0.1014= {\frac{2}{g}}$

$g=19.71=20m/sec^2$

So acceleration due to gravity is 20 $m/sec^2$

So option (E) will be correct answer.

8 0

3 years ago

What magnification would be obtained if an eyepiece with a focal length of 0.38 m was placed on telescope?

weqwewe [10]

Answer:

This question is incomplete

Explanation:

This question is incomplete because the telescope's focal length was not provided. The formula to be used here is

Magnification = telescope's focal length/eyepiece's focal length

The eyepiece's focal length was provided in the question as 0.38 m.

NOTE: Magnification can be described as the power of an instrument (in this case telescope) to enlarge an object. It has no unit and thus the two focal lengths mentioned in the formula above must be in the same unit (preferably meters since one of them is in meters already).

7 0

2 years ago

A 12-pack of Omni-Cola (mass 4.30 kg) is initially at rest ona horizontal floor. It is then pushed in a straight line for1.20 m

erastovalidia [21]

Answer:

a) W = 46.8 J and b) v = 3.84 m/s

Explanation:

The energy work theorem states that the work done on the system is equal to the variation of the kinetic energy

W = ΔK = $k_{f}$ -K₀

a) work is the scalar product of force by distance

W = F . d

Bold indicates vectors. In this case the dog applies a force in the direction of the displacement, so the angle between the force and the displacement is zero, therefore, the scalar product is reduced to the ordinary product.

W = F d cos θ

W = 39.0 1.20 cos 0

W = 46.8 J

b) zero initial kinetic language because the package is stopped

W - $W_{fr}$ = $k_{f}$ -K₀

W - fr d= ½ m v² - 0

W - μ N d = ½ m v

on the horizontal surface using Newton's second law

N-W = 0

N = W = mg

W - μ mg d = ½ m v

v² = (W -μ mg d) 2/m

v = √(W -μ mg d) 2/m

v = √[(46.8 - 0.30 4.30 9.8 1.20) 2/4.3 ]

v = √(31.63 2/4.3)

v = 3.84 m/s

8 0

3 years ago

During an experiment, Ellie records a measurement of 0.0034 m. How would

Goshia [24]

Answer:

(A) She needs to move the decimal point by 3 places

8 0

3 years ago