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2 years ago

Q3, A ball of mass 5.0 kg moving with a Velocity of 10.0 ms collides

Physics

1 answer:

Mrrafil [7]2 years ago

5 0

Answer:

Their common velocity after the collision will be 5.5m/s

Explanation:

look at the attachment above ☝️

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If it takes a ball dropped from rest 2.261 s to fall to the ground, from what height H was it released? Express your answer in m

algol [13]

Answer:

Height, H = 25.04 meters

Explanation:

Initially the ball is at rest, u = 0

Time taken to fall to the ground, t = 2.261 s

Let H is the height from which the ball is released. It can be calculated using the second equation of motion as :

$H=ut+\dfrac{1}{2}at^2$

Here, a = g

$H=\dfrac{1}{2}gt^2$

$H=\dfrac{1}{2}\times 9.8\times (2.261)^2$

H = 25.04 meters

So, the ball is released form a height of 25.04 meters. Hence, this is the required solution.

7 0

2 years ago

Y'all I need this comon please give me a right answer

Natali [406]

It will be a virtual image that appears on the left side of the mirror

i hope this helps!

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3 years ago

An object of length 3.00 cm is inside a plastic block with index of refraction 1.40. If the object is viewed from directly above

Lilit [14]

The length of its image = $-3cm$

Given,

length of object, $h_o = 3 cm$

We know, for flat refracting surface,

Image distance = object distance

So,

magnification is = $-1$

length of the image,

$h_i = magnification * h_o\\\\h_i = -1 * 3 \\\\h_i = -3cm$

Here, negative sign means inverted image.

For more information on refraction, visit

brainly.com/question/14760207?referrer=searchResults

3 0

2 years ago

Wich type of mutation occurs in reproductive cell and can be passed to offspring? A sommatic mutation B germline mutation C poin

ryzh [129]

The answer to your question is,

B. Germline mutation

-Mabel <3

7 0

3 years ago

Capacitances of 10uF and 20uF are connected in parallel,

jarptica [38.1K]

Answer:

The equivalent capacitance will be $15\mu F$

Explanation:

We have given two capacitance $C_=10\mu F\ and\ C_2=20\mu F$

They are connected in parallel

So equivalent capacitance $C=C_1+C_2=10+20=30\mu F$

This equivalent capacitance is now connected in series with $30\mu F$

In series combination of capacitors the equivalent capacitance is given by $\frac{1}{C}=\frac{1}{30}+\frac{1}{30}$

$C=\frac{30}{2}=15\mu F$

So the equivalent capacitance will be $15\mu F$

5 0

3 years ago