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2 years ago

The velocity of a an object in linear motion changes from +25 meters per second to +15 meters per second in 2.0 seconds.

Physics

1 answer:

kodGreya [7K]2 years ago

5 0

Answer:

$-5.0 m/s^2$

Explanation:

Missing question:

What is the object's acceleration?

Solution:

The acceleration of an object is given by

$a=\frac{v-u}{t}$

where

v is the final velocity

u is the initial velocity

t is the time taken for the change in velocity

For the object in this problem,

u = +25 m/s

v = +15 m/s

t = 2.0 s

Substituting,

$a=\frac{15-25}{2}=-5.0 m/s^2$

And the acceleration is negative because its direction is opposite to that of the velocity.

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Onur drops a basketball from a height of 10\,\text{m}10m10, start text, m, end text on Mars, where the acceleration due to gravi

Doss [256]

Answer:

Explanation:

Given that,

Basket ball is drop from height

H=10m

It is dropped on planet mass

And the acceleration due to gravity on Mars is given as

g= 3.7m/s²

Time taken for the ball to reach the ground

Initial velocity of the body is zero

u=0m/s

Using equation of motion: free fall

H = ut + ½gt²

10 = 0•t + ½ × 3.7 ×t²

10 = 0 + 1.85t²

10 = 1.85t²

Then, t² =10/1.85

t² = 5.405

t = √ 5.405

t = 2.325seconds

So the time the ball spend on the air before reaching the ground is 2.325 seconds

5 0

2 years ago

A careful photographic survey of Jupiter’s moon Io by the spacecraft Voyager 1 showed active volcanoes spewing liquid sulfur to

Y_Kistochka [10]

Answer:

529.15 m/s

Explanation:

h = Maximum height = 70000 m

g = Acceleration due to gravity = 2 m/s²

m = Mass of sulfur

As the potential and kinetic energies are conserved

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 2\times 70000}\\\Rightarrow v=529.15\ m/s$

The speed with which the liquid sulfur left the volcano is 529.15 m/s

7 0

2 years ago

What is the tangential velocity at the edge of a disk of radius 10cm when it spins with a frequency of 10Hz? Give your answer wi

Nina [5.8K]

Answer:

630cm/s

Explanation:

In simple harmonic motion, the tangential velocity is expressed mathematically as v = ὦr

ὦ is the angular velocity = 2πf

r is the radius of the disk

f is the frequency

Given the radius of disk = 10cm

frequency = 10Hz

v = 2πfr

v = 2π×10×10

v = 200π

v = 628.32 cm/s

The tangential velocity = 630cm/s ( to 2 significant figures)

8 0

2 years ago

After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of r

mina [271]

Answer:

t = 103.45 n m

Explanation:

given,

refractive index of cornea = 1.38

refractive index of eye drop = 1.45

wavelength of refractive index = 600 nm

refractive index of eye drop is greater than refractive index of cornea and the air.

Formula used in this case

for constructive interference

$2 n t = (m + \dfrac{1}{2})\lambda$

At m = 0 for the minimum thickness, so

$2\times 1.45 \times t = (0 + 0.5)\times 600$

$2.9 \times t =300$

$t =\dfrac{300}{2.9}$

t = 103.45 n m

the minimum thickness of the film of eyedrops t = 103.45 n m

6 0

2 years ago

An older camera has a lens with a focal length of 60mm and uses 34-mm-wide film to record its images. Using this camera, a photo

lesya692 [45]

Answer:

24.71 mm

Explanation:

Distance is proportional to focal length, so

d∝f

which means

$\frac{d'_1}{d'_2}=\frac{f_1}{f_2}$

Magnification of first lens

$M_2=-\frac{d'_1}{d_1}$

and

$M_2=\frac{h'_1}{h_1}$

Similarly, magnification of second lens

$M_2=-\frac{d'_2}{d_1}$

and

$M_2=\frac{h'_2}{h_1}$

From the above equations we get

$\frac{M_1}{M_2}=\frac{d'_1}{d_2'}$

and

$\frac{M_1}{M_2}=\frac{h'_1}{h_2'}$

which means,

$\frac{d'_1}{d_2'}=\frac{h'_1}{h_2'}$

and

$\frac{d'_1}{d_2'}=\frac{f_1}{f_2}$

So, we get

$\frac{f_1}{f_2}=\frac{h'_1}{h_2'}\\\Rightarrow f_2=f_1\times\frac{h_2'}{h'_1}\\\Rightarrow f_2=60\times\frac{14}{34}=24.71\ mm$

∴ Focal length should this camera's lens is 24.71 mm

6 0

2 years ago