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zhenek [66]
3 years ago
5

Which is an example of the force of attraction between two objects that have mass?

Physics
1 answer:
Mariulka [41]3 years ago
4 0

Answer:

Gravity is an example of the force of attraction between two objects that have mass.

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Positively-charged particles consisting of two protons and two neutrons emitted by radioactive materials are
BARSIC [14]

Answer:

The answer to your question is Alpha particles.

Explanation: An electron released by a radioactive nucleus that causes a neutron to change into a proton is called a beta particle.

6 0
3 years ago
Which situation would create a field like the one shown here?
kiruha [24]
D. Interaction between a magnetic South Pole and a copper bar
7 0
3 years ago
A type of cuckoo clock keeps time by having a mass bouncing on a spring, usually something cute like a cherub in a chair. What f
Reil [10]

Answer:0.906 N/m

Explanation:

Given

time period T=0.66 s

mass m=0.01 kg

System can be considered as spring mass system

Time Period of spring mass system is given by

T=2\pi \sqrt{\frac{m}{k}}

squaring

k=\frac{4\pi ^2m}{T^2}

k=\frac{4\times \pi ^2\times 0.01}{0.66^2}

k=0.906 N/m

4 0
3 years ago
A child and sled with a combined mass of 50 kg, start from rest and slide down a frictionless hill that is 7.5 meters high. what
amid [387]
The mechanical enegia is the sum of the kinetic energy plus the potential energy
 Kinetic energy = (1/2) * m * v ^ 2
 Potential energy = m * g * h
 Mechanical energy = (1/2) * m * v ^ 2 + m * g * h
 What is the mechanical energy of the sled at the top? 
 Mechanical energy = (1/2) * m * v ^ 2 + m * g * h
 Mechanical energy = (1/2) * (50) * (0) ^ 2 + (50) * (9.8) * (7.5) = 3675
 Mechanical energy = 3675J
 What is the mechanical energy of the sled at the bottom? 
 By conservation of energy we have that the energy in point 1 is equal to the energy in point 2
 Mechanical energy = 3675J
 What is the speed of the sled at the bottom of the hill?
 Mechanical energy = 3675J = (1/2) * m * v ^ 2
 clearing up v we have
 (1/2) * (50) * v ^ 2 = 3675
 v ^ 2 = 3675 * (2) * (1/50)
 v = root (3675 * (2) * (1/50)) = 12.12 m / s
 answer
 3675J
 3675J
 12.12 m / s
3 0
3 years ago
A net force of 1.6×10−15 N acts on an electron over a displacement of 5.0 cm, in the same direction as the net force. (a) What i
Margaret [11]

Explanation:

It is given that,

Net force acting on the electron, F=1.6\times 10^{-15}\ N

Displacement, d = 5 cm = 0.05 m

(a) Let \Delta E is the change in kinetic energy of the electron. It can be calculated using work energy theorem. Mathematically, it is given by :

W=\Delta E

\Delta E=F\times d

\Delta E=1.6\times 10^{-15}\ N\times 0.05\ m

\Delta E=8\times 10^{-17}\ J

(b) Initial speed of the electron, u = 0

Again using the work energy theorem as :

E=\dfrac{1}{2}m(v^2-u^2)

E=\dfrac{1}{2}m(v^2)

v=\sqrt{\dfrac{2E}{m}}

v=\sqrt{\dfrac{2\times 8\times 10^{-17}}{9.1\times 10^{-31}}}    

v=1.32\times 10^7\ m/s

Hence, this is the required solution.

7 0
3 years ago
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