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lord [1]
3 years ago
10

What does the statement what does the statement 10 m/s to the north describe

Physics
2 answers:
nadezda [96]3 years ago
6 0

Answer:

10 m/s to the north shows the velocity of an object.

Explanation:

The statement ''10 m/s to the north'' shows the velocity of an object. We know that velocity is a vector quantity and speed is scalar.

In the given statement 10 m/s shows the speed of an object while north shows the direction of motion. We can say that,

Speed + direction = velocity

Hence, the given statement shows the velocity of an object.

leva [86]3 years ago
5 0
The statement describes velocity
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A conductor carrying a conventional current in the positive-x direction is in a magnetic field whose vector points in the positi
a_sh-v [17]

Answer:

Positive Z direction (out of screen)

Explanation:

Magnetic force is given by F = il \wedge B. A quick way to gauge the components is to put your left middle finger on the direction of the current, your index on the direction of the magnetic field, and the thumb gives you the answer you want.

4 0
3 years ago
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If i took physical science in 7thgrade do i get credit for physics
Anni [7]

Answer:

I don't think so

Explanation:

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2 years ago
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A hockey puck sliding along frictionless ice with speed v to the right collides with a horizontal spring and compresses it by 1.
patriot [66]

Answer:

The spring's maximum compression will be 2.0 cm

Explanation:

There are two energies in this problem, kinetic energy K= \frac{mv^{2}}{2} and elastic potential energy U= \frac{kx^{2}}{2} (with m the mass, v the velocity, x the compression and k the spring constant. ) so the total mechanical energy at every moment is the sum of the two energies:

E=K+U

Here we have a situation where the total mechanical energy of the system is conserved because there are no dissipative forces (there's no friction), so:

E_{i}=E_{f}

K_{i}+U_{i}=K_{f}+U_{f}

Note that at the initial moment where the hockey puck has not compressed the spring all the energy of the system is kinetic energy, but for a momentary stop all the energy of the system is potential elastic energy, so we have:

K_{i} = U_{f}

\frac{mv^{2}}{2}=\frac{kx^{2}}{2} (1)

Due conservation of energy the equality (1) has to be maintained, so if we let k and m constant x has to increase the same as v to maintain the equality. Therefore, if we increase velocity to 2v we have to increase compression to 2x to conserve the equality. This is  2(1.0) = 2.0 cm

7 0
3 years ago
The current in the wires of a circuit is 60.0 milliAmps. If the voltage impressed across the ends of the circuit were halved (i.
Ksivusya [100]

Answer:

30 miliAmps

Explanation:

Step 1:

Obtaining an expression to solve the question. This is illustrated below:

From ohm's law,

V = IR

Were:

V is the voltage.

I is the current.

R is resistance.

From the question given, we were told that the resistance is constant. Therefore the above equation can be written as shown below:

V = IR

V/I = constant

V1/I1 = V2/I2

V1 is initial voltage.

V2 final voltage.

I1 is initial current.

I2 final current.

Step 2:

Data obtained from the question. This include the following:

Initial voltage (V1) = V

Initial current (I1) = 60 miliAmps

Final voltage (V2) = one-half of the original voltage = 1/2V = V/2

Final current (I2) =..?

Step 3:

Determination of the new current. This can be obtained as follow:

V1/I1 = V2/I2

V/60 = (V/2) / I2

Cross multiply to express in linear form

V x I2 = V/2 x 60

V x I2 = V x 30

Divide both side by V

I2 = (V x 30)/V

I2 = 30mA.

Therefore, the new current is 30miliAmps

5 0
4 years ago
An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.285 T. If the kinetic energy of the electr
Elis [28]

Answer:

The speed of the electron is 6.79 x 10⁵ m/s

The radius of the circular path is 1.357 x 10⁻⁵ m

Explanation:

Given;

magnetic field, B = 0.285 T

energy of electron, E = 2.10 x 10⁻¹⁹ J

The kinetic energy of the electron is calculated as;

K.E = \frac{1}{2} m_eV^2

Where;

m_e is the mass of electron = 9.11 x 10⁻³¹ kg

V is the speed of the electron

K.E = \frac{1}{2} m_eV^2\\\\V^2 = \frac{2.K.E}{m_e} \\\\V = \sqrt{\frac{2K.E}{m_e} } \\\\V = \sqrt{\frac{2*(2.1*10^{-19})}{9.11*10^{-31}} }\\\\V = 6.79 *10^{5} \ m/s

The radius of the circular path is given by;

R = \frac{M_eV}{qB}

where;

q is the charge of the electron = 1.6 x 10⁻¹⁹ C

R = \frac{M_eV}{qB} \\\\R = \frac{9.11 *10^{-31}*6.79 *10^{5}}{1.6*10^{-19}*0.285} \\\\R = 1.357 *10^{-5} \ m

8 0
4 years ago
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