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jeka57 [31]
4 years ago
12

A box is being pushed across the floor at a constant velocity with an

Physics
1 answer:
Oksana_A [137]4 years ago
8 0

Answer:

<em>Magnitude of frictional force = 30 N</em>

Explanation:

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

Fn=m.a

The net force is the vector addition of each individual force. If the sum of all the forces acting on an object is zero, then the acceleration is zero. That means the object moves at a constant speed or is at rest.

When an object is pushed across a horizontal rough surface, there are two forces acting in the direction of the motion: The applied force and the frictional force.

If the applied force is greater than the frictional force, then the object moves at a constant positive acceleration. If the frictional force is greater than the applied force, then the object won't move at all (if it was at rest) or will start a deaccelerated motion (braking).

Finally, if both forces are equal, the object will move at a constant speed or remains at rest. Since the box is moving at a constant speed, we can conclude the frictional force equals the applied force:

Magnitude of frictional force = 30 N

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What is 6.39 times 10 to the 23rd power
Sveta_85 [38]

Answer:

6.39*(10 to the 23rd)=

6.39*10²³

6 0
3 years ago
A polar bear runs at a speed of 11 m/s and has a mass of 380.2 kg. How much Kinetic energy does the bear have?
Yanka [14]

Answer:

\boxed{\sf Kinetic \ energy \ of \ the \ bear (KE) = 23002.1 \ J}

Given:

Mass of the polar bear (m) = 6.8 kg

Speed of the polar bear (v) = 5.0 m/s

To Find:

Kinetic energy of the polar bear (KE)

Explanation:

Formula:

\boxed{ \bold{\sf KE =  \frac{1}{2} m {v}^{2} }}

Substituting values of m & v in the equation:

\sf \implies KE =  \frac{1}{2}  \times 380.2 \times  {11}^{2}

\sf \implies KE = \frac{1}{ \cancel{2}}  \times  \cancel{2} \times 190.1 \times 121

\sf \implies KE = 190.1 \times 121

\sf \implies KE = 23002.1 \: J

\therefore

Kinetic energy of the polar bear (KE) = 23002.1 J

5 0
3 years ago
A toroid has a square cross section with the length of an edge equal to the radius of the inner surface. The ratio of the magnit
Tasya [4]

Answer:

2

Explanation:

7 0
3 years ago
Read 2 more answers
The grounded conductor and the grounding conductor are one and the same conductor and perform the same function.
STALIN [3.7K]

Answer:

Grounded  used to carry the current in the normal conditions. while on the other side grounding is refer to that safety wire that is connected to the earth and currently does not transmit through it.

As grounding wire is referred to as safety wire hence it carries current only in an emergency like a short circuit.

Explanation:

A grounded conductor is referred to as one of the wire that needed in an electric circuit. it is basically a neutral conductor.  It used to carry the current in the normal conditions. while on the other side grounding is refer to that safety wire that is connected to the earth and currently does not transmit through it.

As grounding wire is referred to as safety wire hence it carries current only in an emergency like a short circuit.

5 0
3 years ago
Particle A of charge 2.70 10-4 C is at the origin, particle B of charge -6.36 10-4 C is at (4.00 m, 0), and particle C of charge
Serhud [2]

Answer:

FC vector representation

F_{C} =(19.03i+13.78j)N

Magnitude of FC

F_{C}=23.495N

Vector direction FC

\beta=35.91 degrees: angle that forms FC with the horizontal

Explanation:

Conceptual analysis

Because the particle C is close to two other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.

The directions of the individual forces exerted by qA and qB on qC are shown in the attached figure; The force (FAC) of qA over qC is repulsive because they have equal signs and the force (FBC) of qB over qC is attractive because they have opposite signs.

The FAC force is up in the positive direction and the FBC force forms an α angle with respect to the x axis.

\alpha = tan^{-1}(\frac{3}{4}) = 36.86 degrees

To calculate the magnitudes of the forces we apply Coulomb's law:

F_{AC} = \frac{k*q_{A}*q_{C}}{r_{AC}^2} Equation (1): Magnitude of the electric force of the charge qA over the charge qC

F_{BC} = \frac{k*q_{B}*q_{C}}{r_{BC}^2} Equation (2) : Magnitude of the electric force of the charge qB over the charge qC

Known data

k=8.99*10^9 \frac{N*m^2}{C^2}

q_{A}=2.70*10^{-4} C

q_{B}=-6.36*10^{-4} C

q_{C}=1.04*10^{-4} C

r_{AC} =3

r_{BC}=\sqrt{4^2+3^2} = 5

Problem development

In the equations (1) and (2) to calculate FAC Y FBC:

F_{AC} =8.99*10^9*(2.70*10^{-4}* 1.04*10^{-4})/(3)^2=28.05N

F_{BC} =8.99*10^9*(6.36*10^{-4}* 1.04*10^{-4})/(5)^2=23.785N

Components of the FBC force at x and y:

F_{BCx}=23.785 *Cos(36.86)=19.03N

F_{BCy}=23.785 *Sin(36.86)=14.27N

Components of the resulting force acting on qC:

F_{Cx} = F_{ACx}+ F_{BCx}=0+19.03=19.03N

F_{Cy} = F_{ACy}+ F_{BCy}=28.05-14.27=13.78N

FC vector representation

F_{C} =(19.03i+13.78j)N

Magnitude of FC

F_{C}= \sqrt{19.03^2+13.78^2} =23.495N

Vector direction FC

\beta = tan^{-1} (\frac{13.78}{19.03})=35.91 degrees: angle that forms FC with the horizontal

7 0
3 years ago
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