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3 years ago

What did j.J. Thomson discover about the composition of atoms?

Physics

1 answer:

nordsb [41]3 years ago

5 0

Answer:

J.J. Thomson discovered the electron by experimenting with a Crookes, or cathode ray, tube. He demonstrated that cathode rays were negatively charged. In addition, he also studied positively charged particles in neon gas.

Explanation:

the answer is the underlined part of the answer section. I hoped this helped you a lot!! Study hard for whatever you are doing!!

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1. The more velocity an object has the harder it is to slow it down, speed it up, or turn it.

deff fn [24]

Answer:

The more velocity an object has the harder it is to slow it down

Explanation:

slow it down

8 0

3 years ago

What is the acceleration of a rock is falling 5000m from a. Cliff in 31.9 seconds?

Alja [10]

Answer:

jumping from cliff sounds nice

Explanation:

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3 years ago

(10%) Problem 10: A 7.25-kg bowling ball moving at 9.85 m/s collides with a 0.875-kg bowling pin, which is scattered at an angle

siniylev [52]

Answer

given,

mass of bowling ball = 7.25 Kg

moving speed of the bowling ball = 9.85 m/s

mass of bowling in = 0.875 Kg

scattered at an angle = θ = 21.5°

speed after the collision = 10.5 m/s

angle of the bowling ball

$tan \theta_1 = \dfrac{-[m_2v_2Sin \theta_2]}{m_1v_1 - (m_2v_2cos \theta_2)}$

$tan \theta_1 = \dfrac{-[0.875\times 10.5 \times Sin 21.5^0]}{7.25\times 9.85 - (0.875\times 10.5 \times cos 21.5^0)}$

$tan \theta_1 = \dfrac{-[3.3672]}{62.86}$

$tan \theta_1 = 0.0536$

$\theta_1 =-3.066^0$

b) magnitude of final velocity

$v = \dfrac{-m_2v_2sin\theta_2}{m_1 sin\theta_1}$

$v = \dfrac{-0.875 \times 10.5 sin21.5^0}{7.25 sin(-3.066^0)}$

v = 8.68 m/s

5 0

3 years ago

A object of mass 3.00 kg is subject to a force Fx that varies with position as in the figure below. A coordinate plane has a hor

Leno4ka [110]

Answer:

Explanation:

An impulse results in a change of momentum.

The impulse is the product of a force and a distance. This will be represented by the area under the curve

a) W = ½(4.00)(3.00) = 6.00 J

b) W = (11.0 - 4.00)(3.00) = 21.0 J

c) W = ½(17.0 - 11.0)(3.00) = 9.00 J

d) ASSUMING the speed at x = 0 is in the direction of applied force

½(3.00)(v₄²) = ½(3.00)(0.450²) + 6.00

v₄ = 2.05 m/s

½(3.00)(v₁₇²) = ½(3.00)(0.450²) + 6.00 + 21.0 + 9.00

v₁₇ = 4.92 m/s

If the initial speed is NOT in the direction of applied force, the final speed will be slightly less in both cases.

7 0

3 years ago

A heat pump has a coefficient of performance that is 60% of the Carnot heat pump coefficient of performance. The heat pump is us

Nataliya [291]

Answer:

$T_C$ =118.8 K= 154.2°C

Explanation:

COP_max of carnot heat pump= $\frac{T_{H} }{T_{H}-T_{C} }$

where T_H and T_C are temperatures of hot and cold reservoirs

Also COP= $\frac{Q_H}{W}$

in the question $COP= \frac{60}{100} \times COP_{max}$

⇒ $\frac{Q_H}{W} =\frac{60}{100}\times\frac{T_H}{T_H-T_C}$

heat is added directly to be as efficient as via heat pump

$Q_H= W$

and T_H= 24° C= 297 K

$1=\frac{60}{100}\times \frac{297}{297-T_C}$

on calculating the above equation we get

$T_C$ =118.8 K

the outdoor temperature for efficient addition of heat to interior of home

$T_C$ =118.8 K= 154.2°C

6 0

3 years ago