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3 years ago

The velocity of a body is increases from 10 m/s ti 15 m/s in 5seconds calculate its acceleration

Physics

2 answers:

Veseljchak [2.6K]3 years ago

7 0

Answer:

acceleration = v-u/ t

= 15-10/5

= 5/5

= 1 m/s2

Explanation:

hope this helped you.

Romashka-Z-Leto [24]3 years ago

6 0

Answer:

$\boxed {\boxed {\sf 1 \ m/s^2}}$

Explanation:

Acceleration is the change in velocity over the change in time. Therefore, the formula for calculating acceleration is:

$a= \frac{v_f-v_i}{t}$

Since the body's velocity increased from 10 meters per second to 15 meters per second 15 m/s is the final velocity and 10 m/s is the initial velocity. The time is 5 seconds.

$v_f$ = 15 m/s

$v_i$ = 10 m/s

t= 5 s

$a= \frac{ 15 \ m/s - 10 \ m/s}{5 \ s}$

Solve the numerator.

15 m/s - 10 m/s = 5 m/s

$a= \frac{ 5 \ m/s }{5 \ s}$

Divide.

$a= 1 \ m/s/s$

$a= 1 m/s^2$

The acceleration is 1 meter per second squared.

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An airplane flies in a loop (a circular path in a vertical plane) of radius 160 m . The pilot's head always points toward the ce

notka56 [123]

Answer:

a) 39.6 m/s b) 4123 N

Explanation:

a) At the top of the loop, all of the forces point downwards (force of gravity and normal force).

Fnet=ma

ma=m(v^2/R) (centripetal acceleration)

mg=m(v^2/R)

m cancels out (this is why pilot feels weightless) so,

g=(v^2/R)

9.8 m/s^2 = v^2/160 m

v^2=1568 m^2/s^2

v=39.6 m/s

b) At the bottom of the loop, the normal force and the force of gravity point in opposite directions. The normal force is the weight felt.

Convert 300 km/hr to m/s

300 km/hr=83.3 m/s

Convert pilot's weight into mass:

760 N = 77.55 kg

Fnet=ma

n-mg=m(v^2/R)

n=(77.55 kg)(((83.3 m/s)^2)/160 m)+(77.55 kg)(9.8 m/s^2)

n=3363.2 N+760 N=4123 N

5 0

3 years ago

An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be

lord [1]

Answer:

the quarterback will be moving back at speed of 0.080625 m/s

the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

Explanation:

Given the data in the question;

How fast will he be moving backward just after releasing the ball?

using conservation of momentum;

m₁v₁ = m₂v₂

v₂ = m₁v₁ / m₂

where m₁ is initial mass ( 0.43 kg )

m₂ is the final mass ( 80 kg )

v₁ is the initial velocity ( 15 m/s )

v₂ is the final velocity

so we substitute

v₂ = ( 0.43 × 15 ) / 80

v₂ = 6.45 / 80

v₂ = 0.080625 m/s

Therefore, the quarterback will be moving back at speed of 0.080625 m/s

b) Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?

we make use of the relation between time, distance and speed;

s = d/t

d = st

where s is the speed ( 0.080625 m/s )

t is time ( 0.30 s )

so we substitute

d = 0.080625 × 0.30

d = 0.0241875 m or 2.41875 cm

Therefore, the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

5 0

3 years ago

Do you think an observer on another planet in the solar system might see eclipses

artcher [175]

Answer:

Yes

Explanation:

Eclipses: Eclipses are also known as game of shadows where one object comes between the star(light source) and another object in a straight line such that the shadow of one object falls on other object. This can occur when the apparent size of the star and the object is almost same.

Talking about the Earth, the geometry is such that the Moon and the Sun are of same apparent size as seen from the Earth. Thus Lunar and Solar eclipse can be seen from the Earth. If we were to go on any other planet the same phenomenon can be seen provided the apparent size of moon and the Sun from that planet is same.

We have seen and recorded many such eclipses on Jupiter. These are from the perspective of Earth. When the moons of Jupiter comes exactly between the Sun and Jupiter the shadow of moon will fall on Jupiter. The places where the shadow falls, one will see a solar eclipse.

8 0

3 years ago

Your companion on a train ride through Illinois notices that telephone poles near the tracks appear to be passing by very quickl

Nikitich [7]

Answer: Relative motion

Explanation: If two objects are moving either towards or away from each other with both having their velocities in a reference frame and someone is outside this reference frame seeing the motion of the two objects.

The observer ( in his own frame of reference) will measure a different velocity as opposed to the velocities of the two object in their own reference frame. p

Both the velocity measured by the observer in his own reference frame and the velocity of both object in their reference is correct.

Velocities of this nature that have varying values based on motion referenced to another body is known as relative velocity.

Motion of this nature is known as relative motion.

Note that the word reference frame is simply any where the motion is occurring and the specified laws of motion is valid

For this example of ours, the reference frame of the companion is the train and the telephone poles has their reference frame as the earth.

The companion will measure the velocity of the telephone poles relative to him and the velocity of the telephone pole relative to an observer outside the train will be of a different value.

6 0

3 years ago

player kicks a soccer ball from ground level and sends it flying at an angle of 30 degrees at a speed of 26 m/s. What is the hor

seropon [69]

Answer

22.5 m/s

Explanation

We shall use the trigonometric ratio cosine to find the horizontal component.

cos = adjacent/hypotenuse

Adjacent is the horizontal and hypotenuse is the fly speed.

cos 30° = horizontal / 26

horizontal velocity = 26 × cos 30°

= 26 × 0.866

= 22.5166

= 22.5 m/s

7 0

3 years ago

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The velocity of a body is increases from 10 m/s ti 15 m/s in 5seconds calculate its acceleration​

The velocity of a body is increases from 10 m/s ti 15 m/s in 5seconds calculate its acceleration