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Kruka [31]
3 years ago
13

PLEASE NEED HELP What is the net force acting on the race car in the picture: Question 1 options: 10 N to the right 3 N to the l

eft 17 N 3 N to the right

Physics
1 answer:
Nezavi [6.7K]3 years ago
7 0

Use Net Force Equation.

Fnet = F1 + F2 + F3 + F4

In which the following main forces in play are:

F1 = Gravitational Force

F2 = Normal Force

F3 = Thrust Force

F4 = Air resistance

Since this is assumed to happen on a flat surface, the gravitational and normal forces cancel out. So we are left with:

Fnet = F(thrust) + F(air resistance)

Now we can use this equation to solve for the net force.

Fnet = 10N [right] + 7N [left]

Fnet = 10N [right] - 7N [right]

Fnet = 3N [right].

Therefore, the total net force acting on this car would be 3N [right]

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A current of 15.0 A flows through an electric heater operating on 120 V. Compute the power consumption of the electric heater.
Rzqust [24]

Answer:

There will be 1800 W power consumption in heater

Explanation:

We have given current flowing in the heater I = 15 A

Voltage on which heater is operating V = 120 volt

We have to find the power consumption in the heater

We know that power consumption is given by P = VI

So power consumption in heater = 120 × 15 = 1800 W

So there will be 1800 W power consumption in heater

5 0
3 years ago
A volleyball player bumps a ball across a net with the velocity and angle shown below. What is the maximum height of the ball?
Marrrta [24]

Answer:

D. 12.4 m

Explanation:

Given that,

The initial velocity of the ball, u = 18 m/s

The angle at which the ball is projected, θ = 60°

The maximum height of the ball is given by the formula

                             h = u² sin²θ/2g  m

Where,

                           g - acceleration due to gravity. (9.8 m/s)

Substituting the values in the above equation

                            h = 18² · sin²60 / 2 x 9.8

                               = 18² x 0.75 / 2 x 9.8

                               = 12.4 m

Hence, the maximum height of the ball attained, h = 12.4 m

6 0
3 years ago
Two parallel-plate capacitors have the same plate area. Capacitor 1 has a plate separation twice that of capacitor 2, and the qu
Luba_88 [7]

Answer:

V_1=8 V_2

Explanation:

Given that:

  • Area of the plate of capacitor 1= Area of the plate of capacitor 2=A
  • separation distance of capacitor 2, d_2=d
  • separation distance of capacitor 1, d_1=2d
  • quantity of charge on capacitor 2, Q_2=Q
  • quantity of charge on capacitor 1, Q_1=4Q

We know that the Capacitance of a parallel plate capacitor is directly proportional to the area and inversely proportional to the distance of separation.

Mathematically given as:

C=\frac{k.\epsilon_0.A}{d}.....................................(1)

where:

k = relative permittivity of the dielectric material between the plates= 1 for air

\epsilon_0 = 8.85\times 10^{-12}\,F.m^{-1}

From eq. (1)

For capacitor 2:

C_2=\frac{k.\epsilon_0.A}{d}

For capacitor 1:

C_1=\frac{k.\epsilon_0.A}{2d}

C_1=\frac{1}{2} [ \frac{k.\epsilon_0.A}{d}]

We know, potential differences across a capacitor is given by:

V=\frac{Q}{C}..........................................(2)

where, Q = charge on the capacitor plates.

for capacitor 2:

V_2=\frac{Q}{\frac{k.\epsilon_0.A}{d}}

V_2=\frac{Q.d}{k.\epsilon_0.A}

& for capacitor 1:

V_1=\frac{4Q}{\frac{k.\epsilon_0.A}{2d}}

V_1=\frac{4Q\times 2d}{k.\epsilon_0.A}

V_1=8\times [\frac{Q.d}{k.\epsilon_0.A}]

V_1=8 V_2

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I think the answer is C
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