Ask question

Ask question

All categories

3 years ago

12

A boy who exerts a 300-N force on the ice of a skating rink is pulled by his friend with a force of 75 N, causing the boy to acc

elerate across the ice. If drag and the friction from the ice apply a force of 5 N on the boy, what is the magnitude of the net force acting on him? (1 point)
A) 380 N
B) 80 N
C) 70 N
D) 370 N

2 answers:

Igoryamba3 years ago

4 0

Answer:

C

Explanation:

Jlenok [28]3 years ago

3 0

Answer:

a is the answer

Explanation:

You might be interested in

Current can be made to flow in a stationary conductor by ............................

lara31 [8.8K]

Answer:

A is the answer. Im only 12 and i hope this explanation helps you.

Explanation:

Lenz's Law of Electromagnetic Induction. Faraday's Law tells us that inducing a voltage into a conductor can be done by either passing it through a magnetic field, or by moving the magnetic field past the conductor and that if this conductor is part of a closed circuit, an electric current will flow.

3 0

3 years ago

Select the correct answer.

Serhud [2]

Answer:

b hope this helps

Explanation:

5 0

3 years ago

Read 2 more answers

*WILL MARK BRAINLIEST FOR RIGHT ANSWER* How much current must be applied across a 60 Ω light bulb filament in order for it to co

sp2606 [1]

Answer: The current must be equal to $\frac{\sqrt{33} }{6}$ amps, or ~0.9574 amps.

Explanation:

You can find the current in amperes using ohms and watts from this formula:

$I = \sqrt{\frac{P}{R} }$

Where P represents power in watts, R represents resistance in ohms, and I represents current in amperes.

You can then substitute 60 and 55 into the equation to find I:

$I = \sqrt{\frac{55}{60} } \\I = \frac{\sqrt{55} }{\sqrt{60} }$

Then, simplify the denominator:

$I = \frac{\sqrt{55} }{2\sqrt{15} }$

Rationalize the denominator:

$I = \frac{\sqrt{55} }{2\sqrt{15} } * \frac{\sqrt{15} }{\sqrt{15} } = \frac{\sqrt{825} }{30}$

Simplify the numerator by finding its factors:

$I = \frac{5\sqrt{33} }{30} = \frac{\sqrt{33} }{6}$

The current must be equal to $\frac{\sqrt{33} }{6}$ amps, or ~0.9574 amps.

8 0

3 years ago

To win the game, a placekicker must kick a football from a point 44 m (48.1184 yd) from the goal, and the ball must clear the cr

Ganezh [65]

Answer:

The distance by the ball clear the crossbar is 1.15 m

Explanation:

Given that,

Distance = 44 m

Speed = 24 m/s

Angle = 31°

Height = 3.05 m

We need to calculate the horizontal velocity

Using formula of horizontal velocity

$u_{x}=u\cos\theta$

Put the value into the formula

$u_{x}=24\cos(31)$

$u_{x}=20.5\ m/s$

We need to calculate the vertical velocity

Using formula of vertical velocity

$u_{y}=u\sin\theta$

Put the value into the formula

$u_{y}=24\sin(31)$

$u_{y}=12.3\ m/s$

We need to calculate the time

Using formula of time

$t=\dfrac{d}{u_{x}}$

Put the value into the formula

$t=\dfrac{44}{20.5}$

$t=2.1\ sec$

We need to calculate the vertical height

Using equation of motion

$h=u_{y}t+\dfrac{1}{2}at^2$

Put the value into the formula

$h=12.3\times2.1-\dfrac{1}{2}\times9.8\times(2.1)^2$

$h=4.2\ m$

We need to calculate the distance by the ball clear the crossbar

Using formula for vertical distance

$d=h-3.05$

Put the value of h

$d=4.2-3.05$

$d=1.15\ m$

Hence, The distance by the ball clear the crossbar is 1.15 m

7 0

3 years ago

A physics student of mass 43.0 kg is standing at the edge of the flat roof of a building, 12.0 m above the sidewalk. An unfriend

Dmitry_Shevchenko [17]

Answer:

The speed of the student just before she lands, v₂ is approximately 8.225 m/s

Explanation:

The given parameters are;

The mass of the physic student, m = 43.0 kg

The height at which the student is standing, h = 12.0 m

The radius of the wheel, r = 0.300 m

The moment of inertia of the wheel, I = 9.60 kg·m²

The initial potential energy of the female student, P.E.₁ = m·g·h₁

Where;

m = 43.0 kg

g = The acceleration due to gravity ≈ 9.81 m/s²

h = 12.0 h

∴ P.E.₁ = 43 kg × 9.81 m/s² × 12.0 m = 5061.96 J

The kinetic rotational energy of the wheel and kinetic energy of the student supporting herself from the rope she grabs and steps off the roof, K₁, is given as follows;

$K_1 = \dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2$

The initial kinetic energy, 1/2·m·v₁² and the initial kinetic rotational energy, 1/2·m·ω₁² are 0

∴ K₁ = 0 + 0 = 0

The final potential energy of the student when lands. P.E.₂ = m·g·h₂ = 0

Where;

h₂ = 0 m

The final kinetic energy, K₂, of the wheel and student is give as follows;

$K_2 = \dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2$

Where;

v₂ = The speed of the student just before she lands

ω₂ = The angular velocity of the wheel just before she lands

By the conservation of energy, we have;

P.E.₁ + K₁ = P.E.₂ + K₂

∴ m·g·h₁ + $\dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2$ = m·g·h₂ + $\dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2$

Where;

ω₂ = v₂/r

∴ 5061.96 J + 0 = 0 + $\dfrac{1}{2} \times 43.0 \, kg \times v_{2}^2+\dfrac{1}{2} \times 9.60 \, kg\cdot m^2 \cdot \left (\dfrac{v_2}{0.300 \, m} }\right ) ^2$

5,061.96 J = 21.5 kg × v₂² + 53. $\overline 3$ kg × v₂² = 21.5 kg × v₂² + 160/3 kg × v₂²

v₂² = 5,061.96 J/(21.5 kg + 160/3 kg) ≈ 67.643118 m²/s²

v₂ ≈ √(67.643118 m²/s²) ≈ 8.22454363 m/s

The speed of the student just before she lands, v₂ ≈ 8.225 m/s.

5 0

3 years ago