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3 years ago

Rutherford's atomic theory included which idea?

Physics

1 answer:

Murljashka [212]3 years ago

4 0

Your answer would be C Atoms are composed of a positively charged sphere that contains electrons.

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A small, 300 g cart is moving at 1.20 m/s on an air track when it collides with a larger, 2.00 kg cart at rest?

stiv31 [10]

Answer:

The speed of the large cart after collision is 0.301 m/s.

Explanation:

Given that,

Mass of the cart, $m_1 = 300\ g = 0.3\ kg$

Initial speed of the cart, $u_1=1.2\ m/s$

Mass of the larger cart, $m_2 = 2\ kg$

Initial speed of the larger cart, $u_2=0$

After the collision,

Final speed of the smaller cart, $v_1=-0.81\ m/s$ (as its recolis)

To find,

The speed of the large cart after collision.

Solution,

Let $v_2$ is the speed of the large cart after collision. It can be calculated using conservation of momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

$m_1u_1+m_2u_2-m_1v_1=m_2v_2$

$v_2=\dfrac{m_1u_1+m_2u_2-m_1v_1}{m_2}$

$v_2=\dfrac{0.3\times 1.2+0-0.3\times (-0.81)}{2}$

$v_2=0.301\ m/s$

So, the speed of the large cart after collision is 0.301 m/s.

4 0

3 years ago

A piston releases 18 J of heat into its surroundings while expanding from 0.0002 m^3 to 0.0006 m^3 at a constant pressure of 1.0

Elodia [21]

Answer:

value of heat is 18 J

2. step by step

formular w=p(volume1-volume2)

w= 1.0×10^5(0.0006-0.0004)

w= 40 J

8 0

3 years ago

A car and a motorcycle start from rest at the same time on a straight track, but the motorcycle is 25.0 m behind the car. The ca

sergeinik [125]

Answer:

t = 8.45 sec

car distance d = 132.09 m

bike distance d = 157.08 m

Explanation:

GIVEN :

motorcycle is 25 m behind the car , therefore distance need to covered by bike to overtake car is 25+ d, when car reache distance d at time t

for car

by equation of motion

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

u = 0 starting from rest

$d = \frac{1}{2}at^2$

$t^2 = \frac{2d}{a}$

for bike

$d+25 = 0 + \frac{1}{2}*4.40t^2$

$t^2= \frac{d+25}{2.20}$

equating time of both

$\frac{2d}{a} = \frac{d+25}{2.20}$

solving for d we get

d = 132 m

therefore t is $= \sqrt{\frac{2d}{a}}$

$t = \sqrt{\frac{2*132}{3.70}}$

t = 8.45 sec

each travelled in time 8.45 sec as

for car

$d = \frac{1}{2}*3.70 *8.45^2$

d = 132.09 m

fro bike

$d = \frac{1}{2}*4.40 *8.45^2$

d = 157.08 m

7 0

3 years ago

If anyone knows please anwser

Nonamiya [84]

Answer:

Explanation:

4 0

3 years ago

Read 2 more answers

You do 30 J of work to load a toy dart gun. However, the dart is 10 cm long and feels a frictional force of 10 N while going thr

shepuryov [24]

Answer: Option (b) is the correct answer.

Explanation:

The given data is as follows.

$K.E_{1}$ = 30 J, l = 10 cm = 0.1 m (as 1 m = 100 cm)

F = 10 N

The formula to calculate the work done is as follows.

W = $F \times l$

Hence, putting the given values into the above formula as follows.

W = $F \times l$

= $10 N \times 0.1 m$

= 1 J

As the dart loses 1 joule of energy when it goes through the barrel. Therefore, the kinetic energy it has when it comes out is calculated as follows.

K.E = 30 - 1

= 29 J

Thus, we can conclude that the kinetic energy of the fired dart is 29 J.

6 0

3 years ago