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polet [3.4K]
3 years ago
15

One terminal of a car battery is said to be connected to “ground.” since it is not really connected to the ground, what is meant

by this expression
Physics
1 answer:
GalinKa [24]3 years ago
3 0
This expression means that the negative terminal (-) is connected to the metal chassis or engine, which means that all voltages used for the electrical devices in the car are measured with respect to the car's chassis or engine. 
Today's vehicles have a negative ground system, which means that the vehicle's steel frame or chassis is directly connected to the negative side of the battery via the negative battery cable. 
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In the photoelectric effect, a photon with an energy of 5.3 × 10–19 J strikes an electron in a metal. Of this energy, 3.6 × 10–1
valentina_108 [34]

Answer:

The velocity of the photo electron is 6.11\times 10^5\ m/s.

Explanation:

Given that,

Supplied energy, E_s=5.3\times 10^{-19}\ J

Minimum energy of the electron to escape from the metal, E_e=3.6\times 10^{-19}\ J

We need to find the velocity of the photo electron. The energy supplied by the photon is equal to the sum of minimum escape energy and the kinetic energy of the escaping electron. So,

5.3\times 10^{-19}\ J=3.6\times 10^{-19}\ J+K\\\\K=5.3\times 10^{-19}-3.6\times 10^{-19}\\\\K=1.7\times 10^{-19}\ J

The formula of kinetic energy is given by :

K=\dfrac{1}{2}mv^2\\\\v=\sqrt{\dfrac{2K}{m}} \\\\v=\sqrt{\dfrac{2\times 1.7\times 10^{-19}}{9.1\times 10^{-31}}} \\\\v=6.11\times 10^5\ m/s

So, the velocity of the photo electron is 6.11\times 10^5\ m/s.

4 0
3 years ago
A beryllium-9 ion has a positive charge that is double the charge of a proton, and a mass of 1.50 ✕ 10−26 kg. At a particular in
seropon [69]

Answer:

Magnetic force, F = 3.52\times 10^{-13}\ N

Explanation:

Given that,

A beryllium-9 ion has a positive charge that is double the charge of a proton, q=2\times 1.6\times 10^{-19}\ C=3.2\times 10^{-19}\ C

Speed of the ion in the magnetic field, v=5\times 10^6\ m/s

Its velocity makes an angle of 61° with the direction of the magnetic field at the ion's location.

The magnitude of the field is 0.220 T.

We need to find the magnitude of the magnetic force on the ion. It is given by :

F=qvB\\\\F=3.2\times 10^{-19}\times 5\times 10^6\times 0.22\\\\F=3.52\times 10^{-13}\ N

So, the magnitude of magnetic force on the ion is 3.52\times 10^{-13}\ N.

3 0
3 years ago
Read 2 more answers
Near the equator, the Earth's magnetic field points almost horizontally to the north and has magnitude B=.5 x 10^-4T. What shoul
Nataly [62]
Base in your question about the magnetic field of the Earth near the equator where as its almost horizontally to the north and has magnitude of B=0.5x10^-4t, the answer is <span>Velocity of electron will be westwards.</span>
6 0
3 years ago
Drawing a shows a displacement vector (450.0 m along the y axis). In this x, y coordinate system the scalar components are Ax 0
Alisiya [41]

Answer:

x ’= 368.61 m,  y ’= 258.11 m

Explanation:

To solve this problem we must find the projections of the point on the new vectors of the rotated system  θ = 35º

            x’= R cos 35

            y’= R sin 35

           

The modulus vector can be found using the Pythagorean theorem

            R² = x² + y²

            R = 450 m

we calculate

            x ’= 450 cos 35

            x ’= 368.61 m

            y ’= 450 sin 35

            y ’= 258.11 m

4 0
3 years ago
5.0 kg, with a bullet of mass 0.1 kg. The target was mounted on
kari74 [83]

Answer:

306 m/s

Explanation:

Law of conservation of momentum

m1v1 + m2v2 = (m1+m2)vf

m1 is the bullet's mass so it is 0.1 kg

v1 is what we're trying to solve

m2 is the target's mass so it is 5.0 kg

v2 is the targets velocity, and since it was stationary, its velocity is zero

vf is the velocity after the target is struck by the bullet, so it is 6.0 m/s

plugging in, we get

(0.1 kg)(v1) + (5.0 kg)(0 m/s) = (0.1 kg + 5.0 kg)(6.0 m/s)

(0.1)(v1) + 0 = 30.6

(0.1)(v1) = 30.6

v1 = 306 m/s

8 0
3 years ago
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