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Katen [24]
4 years ago
12

Which of the following is an example of conduction?

Physics
1 answer:
Jet001 [13]4 years ago
7 0

the answer was C because Heat is transferred by conduction when a hot object is in direct contact with a cold object. So, a metal spoon becoming warm when placed in a cup of hot tea is an example of conduction because the spoon gains heat by being in direct contact with the hot tea.

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A hydrogen atom has its electron in the n = 6 level. the radius of the electron's orbit in the bohr model is 1.905 nm.
gayaneshka [121]
Are there any options or is it not multiple choice.
5 0
3 years ago
What is the force on an electron in a CRT when it’s moving at 2.5 × 105 meters/second perpendicular to a magnetic field of 1.5 t
Firdavs [7]
F = Magnetic Force
B = Magnetic Field
V = Velocity

*The vectors from the photo you get doing the left-hand rule.

The magnetic force is always perpendicular to the magnetic field.

And as told in the statement, the electron is moving perpendicular to a magnetic field, that is, the Velocity forms an 90 degree angle / Right angle with the magnetic field.

The formula to find the Magnetic Force is:

f = |q| \times v \times b \times sin \: \theta

Where "q" is the Charge and the sin theta is the angle formed by the Velocity and Magnetic Field, in this case it's 90°. Sin 90° = 1.

f = |- 1.6 \times {10}^{ - 19} | \times 2.5 \times {10}^{5} \times 1.5 \times 1 \\ f = 6 \times {10}^{ - 19 + 5} \\ f = 6 \times {10}^{ - 14} \: newtons
Newton (N) = C x m/s x T = (C x m x T)/s

4 0
3 years ago
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WINSTONCH [101]
Electromagnetic waves..............................
3 0
3 years ago
An object of unknown mass is hung on the end of an unstretched spring and is released from rest. The acceleration of gravity is
lakkis [162]

Answer:

0.33 s

Explanation:

For this case, as the object is hung on the end of an unstretched spring, we can consider this system as a simple pendulum.

For this system, we can determine the period of the motion using the following formula:

T = 2π√(L/g)

Where: T = period (in sec), L = lenght of the spring, g = acceleration of garvity = 9.8 m/s²

By the exact time the object is 2.75 cm before coming to rest, that will be the lenght of the spring we can consider (2.75 cm = 0.0275 m)

Finally:

T = 2π√(0.00275/9.8)

T = 0.33 sec

4 0
3 years ago
A force F = (2.75 N)i + (7.50 N)j + (6.75 N)k acts on a 2.00 kg mobile object that moves from an initial position of d = (2.75 m
Natasha2012 [34]

Answer:

The work done on the object by the force in the 5.60 s interval is 40.93 J.

Explanation:

Given that,

Force F=(2.75\ N)i+(7.50\ N)j+(6.75\ N)k

Mass of object = 2.00 kg

Initial position d=(2.75)i-(2.00)j+(5.00)k

Final position d=-(5.00)i+(4.50)j+(7.00)k

Time = 4.00 sec

We need to calculate the work done on the object by the force in the 5.60 s interval.

Using formula of work done

W=F\cdot d

W=F\cdot(d_{f}-d_{i})

Put the value into the formula

W=(2.75)i+(7.50)j+(6.75)k\cdot (-(5.00)i+(4.50)j+(7.00)k-(2.75)i+(2.00)j-(5.00)k)

W=(2.75)i+(7.50)j+(6.75)k\cdot((-7.75)i+6.50j+2.00k)

W=2.75\times-7.75+7.50\times6.50+6.75\times2.00

W=40.93\ J

Hence, The work done on the object by the force in the 5.60 s interval is 40.93 J.

4 0
3 years ago
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