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3 years ago

An electron has _____

Physics

2 answers:

Soloha48 [4]3 years ago

8 0

Answer:

option (D)

Explanation:

The elementary particles of atom are protons, neutrons and electrons. The protons and neutrons are residing in the nucleus and electron revolves around the nucleus in fixed stationary orbits.

The mass of proton and a neutron is almost same and equal to 1.67 x 10^-27 Kg.

The mass of an electron is 9.1 x 10^-31 Kg.

The mass of electron is much less than the mass of proton or a neutron.

Proton is 1837 times heavier than the electron.

Marrrta [24]3 years ago

3 0

a proton and neutron weigh about the same, while an electron is about 1/2000 of the weight of either a proton or neutron

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Find the network done by friction on a box that moves in a complete circle of radius 1.82 m on a uniform horizontal floor. The c

m_a_m_a [10]

Answer:

C) W = - 190 J

Explanation:

Notation

Wf = work done by the friction force (unknown)

Ff = force of the friction

d = distance travelled by the box = (2 pi 1.82 m) = 11.435 m

6 0

3 years ago

Read 2 more answers

As you read this in your chair, how fast are you moving relative to the chair? relative to the sun?

Oliga [24]

To answer this question, you need to know the definition of Relative Motion:
The motion is relative when it depends on a reference point or referencial system. If you know the reference point, you can determine the velocity of an object.
If you are sitting on your chair, you are not moving relative to it (Your speed is 0 km/s); but as you know, our planet moves around the Sun (Traslation Movement) with a speed of 30.0 km/s. Therefore, you are moving 30.0 km/s relative to the sun.

5 0

3 years ago

Block b rests upon a smooth surface. if the coefficients of static and kinetic friction between a and b are μs = 0.4 and μk = 0.

aliina [53]

Given

Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient of kinetic friction µk = 0.3. If a force P is applied to the body, no relative motion will take place until the applied force is equal to the force of friction Ff, which is acting opposite to the direction of motion. Magnitude of static force of friction between block A and block B, Fs = µsN, where N is reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA + mB)a,

6 = (20 / 32.2 + 50 / 32.2)a

2.173a = 6

A = 2.76 ft/s^2

To check slipping occurs between block A and block B, consider block A:

P – Ff = mAaA

6 – Ff = 1.71

Ff = 4.29 lb

And also,

N = wA. We know static friction,

Fs = µsN

Fs = 0.4 x 20

Fs = 8lb

Frictional force is less than static friction. Ff < Fs

<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of block B aB = 2.76 ft/s^2</span>

6 0

3 years ago

A wire 95.0 mm long is bent in a right angle such that the wire starts at the origin and goes in a straight line to x = 40.0 mm

qaws [65]

Answer:

0.1040512455 N

$36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction$

0.05925 N

$29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction$

Explanation:

I = Current

B = Magnetic field

Separation between end points is

$l=\sqrt{40^2+55^2}\\\Rightarrow l=68.00735\ mm$

Effective force is given by

$F=IlB\\\Rightarrow F=5.1\times 68.00735\times 10^{-3}\times 0.3\\\Rightarrow F=0.1040512455\ N$

The force is 0.1040512455 N

$tan\theta=\dfrac{55}{40}\\\Rightarrow \theta=tan^{-1}\dfrac{55}{40}\\\Rightarrow \theta=53.97^{\circ}$

The angle the force makes is given by

$\alpha=\theta-90\\\Rightarrow \alpha=53.97-90\\\Rightarrow \alpha=-36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction$

The direction is $36.03^{\circ}\ or\ 323.97^{\circ}in\ CCW\ direction$

$F=IlB\\\Rightarrow F=4.9\times \sqrt{20^2+35^2}\times 10^{-3}\times 0.3\\\Rightarrow F=0.05925\ N$

The force is 0.05925 N

$tan\theta=\dfrac{35}{20}\\\Rightarrow \theta=tan^{-1}\dfrac{35}{20}\\\Rightarrow \theta=60.26^{\circ}$

$\alpha=\theta-90\\\Rightarrow \alpha=60.26-90\\\Rightarrow \alpha=-29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction$

The direction is $29.74^{\circ}\ or\ 330.26^{\circ}in\ CCW\ direction$

4 0

3 years ago

(b) If you decrease the length of the pendulum by 25%, how does the new period TN compare to the old period T?

malfutka [58]

Answer:

The new period will be reduced by 50%

Explanation:

The period of pendulum is given by;

$T= 2\pi\sqrt{\frac{L}{g} }\\\\\frac{T}{2\pi} = \sqrt{\frac{L}{g} }\\\\(\frac{T}{2\pi} )^2 = {\frac{L}{g}}\\\\\frac{T^2}{4\pi ^2} = {\frac{L}{g}}\\\\T^2(\frac{g}{4\pi ^2}) = L\\\\ \frac{g}{4\pi ^2}= \frac{L}{T^2}\\\\\frac{L_1}{T_1^2} = \frac{L_2}{T_2^2}$

When the length is decreased by 25%, the new length L₂ is given by;

L₂ = 25/100(L₁)

L₂ = 0.25L₁

$\frac{L_1}{T_1^2} = \frac{L_2}{T_2^2}\\\\T_2^2 = \frac{T_1^2L_2}{L_1} \\\\T_N^2 = \frac{T^2(0.25L_1)}{L_1}\\\\ T_N^2 =0.25T^2\\\\T_N = \sqrt{0.25T^2}}\\\\T_N = 0.5 T$

Thus, the new period will be reduced by 50%

8 0

3 years ago