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zheka24 [161]
3 years ago
10

A horizontal force of 92.7 N is applied to a 40.5 kg crate on a rough, level surface. If the crate accelerates at 1.13 m/s2, wha

t is the magnitude of the force of kinetic friction (in N) acting on the crate
Physics
1 answer:
lord [1]3 years ago
5 0

Answer:

The value is F_f =  46.935 \  N

Explanation:

From the question we are told that

    The  magnitude of the horizontal force is F  =  92.7 \  N

     The mass of the crate is  m  =  40.5 \  kg

     The acceleration of the crate is  a =  1.13 \ m/s

Generally the net force acting on the crate is mathematically represented as

       F_{net} =  F -  F_f =  ma

Here F_f is force of kinetic friction (in N) acting on the crate

      So  

            92.7  -  F_f =  40.5 * 1.13

=>         F_f =  46.935 \  N

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The magnetic field inside a superconducting solenoid is 4.00 T. The solenoid has an inner diameter of 6.20 cm and a length of 26
Delvig [45]

Answer:

(a) The magnetic energy density in the field is 6.366 J/m³

(b) The energy stored in the magnetic field within the solenoid is 5 kJ

Explanation:

magnitude of magnetic field inside solenoid, B = 4 T

inner diameter of solenoid, d = 6.2 cm

inner radius of the solenoid, r = 3.1 cm = 0.031 m

length of solenoid, L = 26 cm = 0.26 m

(a) The magnetic energy density in the field is given by;

u _B = \frac{B^2}{2\mu_o} \\\\u _B = \frac{(4)^2}{2(4\pi*10^{-7})}\\\\u_B = 6.366*10^6 \ J/m^3

(b) The energy stored in the magnetic field within the solenoid

U_B = u_B V\\\\U_B = u_B AL

U_B = u_B(A)(L)\\\\U_B = 6.366*10^6(\pi * 0.031^2)(0.26) \\\\U_B = 4997.69 J\\\\U_B = 5 \ KJ\\

6 0
3 years ago
PLEASE HELPPPPP
Ganezh [65]

The only graph that accurately depict the given motion is graph D.

The given parameters;

  • initial position of the man = 0
  • direction of the man's first displacement = backward
  • time of first motion, t₁ = 6 seconds
  • velocity of this first displacement = v₁
  • time without any motion (<em>zero movement</em>) = 6 seconds
  • direction of the second displacement = forward
  • velocity of second displacement = 2v₁

Let the acceleration of the first displacement = a

Acceleration of the second displacement = 2a

From the given graphs we can eliminate every graph without initial decrease or motion towards the negative direction.

The only options with initial motion towards the negative direction are;

  • <em>graph B</em>, and
  • <em>graph D</em>.

The difference between graph B and D;

  • in graph B there is a uniform motion for 6 seconds
  • in graph D there is no motion for 6 seconds (<em>this is obvious as the line fall directly on top of the horizontal axis maintaining a value of zero for 6 seconds</em>).

Thus, the only graph that accurately depict the given motion is graph D.

Learn more here: brainly.com/question/21095906

5 0
2 years ago
Can I PLEASE get some help?
Rus_ich [418]
1.) C
2.) B
3.) D
4.) B
Good luck with your work!
3 0
4 years ago
A ball is thrown straight up in the air. For which situations are both the instantaneous velocity and the acceleration zero? (Se
Gwar [14]

Answer:

Therefore, the situation in which both the instantaneous velocity and acceleration become zero, is the situation when the ball reaches the highest point of its motion.

Explanation:

When a ball is thrown upward under the free fall action of gravity, it starts to loose its Kinetic Energy as it moves upward. As the ball moves in upward direction, its kinetic energy gradually converts into its potential energy. As a result the speed of the ball starts to decrease as it moves up. Therefore, at the highest point during its motion, the velocity of ball becomes zero and it stops at the highest point for a moment, and then it starts to fall back down, under the influence of gravitational force.

Therefore, the situation in which both the instantaneous velocity and acceleration become zero, is the situation <u>when the ball reaches the highest point of its motion.</u>

6 0
3 years ago
A 17.0 resistor and a 6.0 resistor are connected in series with a battery. The potential difference across the 6.0 resistor is m
tia_tia [17]

Answer:

V= 57.5 V

Explanation:

  • If the resistors are in the linear zone of operation, the potential difference across them, must obey Ohm's law:

        V = I*R

  • For the 6.0 Ω resistor, if the potential difference across it is 15 V, we can find the current flowing through it as follows:

       I = \frac{V}{R} = \frac{15 V}{6.0 \Omega} = 2.5 A

  • In a series circuit, the current is the same at any point of it, so the current through the battery is I = 2.5 A
  • The equivalent resistance of a series circuit is just the sum of the resistances, so, in this case, we can write the following equation:

      R_{eq} = R_{1} +R_{2} = 17.0 \Omega + 6.0 \Omega = 23.0 \Omega

  • Applying Ohm's Law to the equivalent resistance, we can find the potential difference through it, that must be equal to the potential difference across the battery, as follows:

        V = I* R_{eq}  = 2.5 A * 23.0 \Omega = 57.5 V

8 0
3 years ago
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