Ask question

Ask question

All categories

GenaCL600 [577]

3 years ago

8

A glider is gliding through the air at a height of 416 meters with a speed of 45.2 m/s. The glider

1 answer:

julia-pushkina [17]3 years ago

3 0

There's not enough information to find an answer.

I think the idea here is that in descending (416 - 278) = 138 meters,
the glider gives up some gravitational potential energy, which
becomes kinetic energy at the lower altitude. This is all well and
good, but we can't calculate the difference in potential energy
without knowing the mass of the glider.

You might be interested in

Cara is building a model of the solar system, which includes the Sun. She plans to include a written description to provide deta

Paladinen [302]

Answer:

She should explain that the Sun is made up of gaseous layers that surround an iron core.

6 0

3 years ago

A man runs 1200m on a straight line in 4 min . find his velocity.

luda_lava [24]

Answer:

5m/sec^2

Explanation:

Distance=1200m

Time=4 min

1=60sec

4=4 x 60

=240sec

Velocity=Distance/Time

Velocity=1200/240

Velocity=5m/sec^2

Mark me as brainliest

7 0

3 years ago

Sphere 1 with radius R_1 has positive charge q, Sphere 2 with radius 4.50 R_1 is far from sphere 1 and initially uncharged. The

tia_tia [17]

Answer:

Explanation:

capacitance of sphere 2 will be 4.5 times sphere 1

a ) when spheres are in contact they will have same potential finally . So

V_1 / V_2 = 1

b )

Charge will be distributed in the ratio of their capacity

charge on sphere1 = q x 1 / ( 1 + 4.5 )

= q / 5.5

fraction = 1 / 5.5

c ) charge on sphere 2

= q x 4.5 / 5.5

fraction = 4.5 / 5.5

d ) surface charge density of sphere 1

= q /( 5.5 x A ) where A is surface area

surface charge density of sphere 2

= q x 4.5 /( 5.5 x 4.5² A ) where A is surface area

= q /( 5.5 x 4.5 A )

q_1/q_2 = 4.5

6 0

3 years ago

This time particle A starts from rest and accelerates to the right at 65.5 cm/s

FrozenT [24]

Answer:

t = 4 s

Explanation:

As we know that the particle A starts from Rest with constant acceleration

So the distance moved by the particle in given time "t"

$d = v_i t + \frac{1}{2}at^2$

$d = 0 + \frac{1}{2}(65.5)t^2$

$d_1 = 32.75 t^2 cm$

Now we know that B moves with constant speed so in the same time B will move to another distance

$d_2 = 44 \times t$

now we know that B is already 349 cm down the track

so if A and B will meet after time "t"

then in that case

$d_1 = 349 + d_2$

$32.75 t^2 = 349 + 44 t$

on solving above kinematics equation we have

$t = 4 s$

4 0

3 years ago

Which of the following is an example of a noninfectious disease?

Scorpion4ik [409]

D lung cancer is not infectious

3 0

3 years ago

Read 2 more answers