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2 years ago

Which force is represented by the arrow at

Physics

2 answers:

shutvik [7]2 years ago

7 0

A is pulling the block straight down toward the center of the Earth, no matter what the slope of the plane may be. A is the force of gravity.

The directions of B and C both depend on the slope of the plane.

B is a force that's parallel to the plane, pulling the block UP the plane. B is the force of friction.

C is a force perpendicular to the plane, preventing the block from falling down through the plane. C is the normal force.

PilotLPTM [1.2K]2 years ago

7 0

Answer:

A is pulling the block straight down toward the center of the Earth, no matter what the slope of the plane may be. A is the force of gravity.

The directions of B and C both depend on the slope of the plane.

B is a force that's parallel to the plane, pulling the block UP the plane. B is the force of friction.

C is a force perpendicular to the plane, preventing the block from falling down through the plane. C is the normal force.

Explanation:

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A sailboat sits tied up at the dock.

shutvik [7]

Answer:B

Explanation:i think so im not suree

5 0

3 years ago

Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the

yaroslaw [1]

Answer:

K_a = 8,111 J

Explanation:

This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved

initial instant. Just before dropping the particles

p₀ = 0

final moment

p_f = m_a v_a + m_b v_b

p₀ = p_f

0 = m_a v_a + m_b v_b

tells us that

m_a = 8 m_b

0 = 8 m_b v_a + m_b v_b

v_b = - 8 v_a (1)

indicate that the transfer is complete, therefore the kinematic energy is conserved

starting point

Em₀ = K₀ = 73 J

final point. After separating the body

Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²

K₀ = K_f

73 = ½ m_a (v_a² + v_b² / 8)

we substitute equation 1

73 = ½ m_a (v_a² + 8² v_a² / 8)

73 = ½ m_a (9 v_a²)

73/9 = ½ m_a (v_a²) = K_a

K_a = 8,111 J

3 0

3 years ago

1. An object is projected upward with a velocity of 125 m/s.

grigory [225]

Answer:

A) s = 796.38 m

B) t = 12.742 s

C) T = 25.484 s

Explanation:

A) First of all let's find the time it takes to get to maximum height using Newton's first equation of motion.

v = u + gt

u = 125 m/s

v = 0 m/s

g = 9.81 m/s²

Thus;

0 = 125 - 9.81(t)

g is negative because motion is against gravity. Thus;

9.81t = 125

t = 125/9.81

t = 12.742 s

Max height will be gotten from Newton's 2nd equation of motion;

s = ut + ½gt²

s = (125 × 12.742) + (½ × -9.81 × 12.742²)

s = 1592.75 - 796.37

s = 796.38 m

B) time to reach maximum height is;

t = u/g

t = 125/9.81

t = 12.742 s

C) Total time elapsed is;

T = 2u/g

T = 2 × 125/9.81

T = 25.484 s

4 0

3 years ago

In creating his definition of horsepower, James Watt, the inventor of the steam engine, calculated the power output of a horse o

Studentka2010 [4]

Answer:

Part a)

$F = 182.3 Lb$

Part b)

$P = 0.55 HP$

Explanation:

Diameter of the circle = 24 ft

Diameter = 731.52 cm = 7.3152 m

now the horse complete 144 trips in one hour

so time to complete one trip is given as

$t = \frac{3600}{144} s$

$t = 25 s$

now the speed of the horse is given as

$v = \frac{2\pi r}{t}$

$v = \frac{\pi(7.3152)}{25}$

$v = 0.92 m/s$

Part a)

Now we know that the power is defined as rate of work done

it is given as

$P = F v$

$746 = F(0.92)$

$F = 810.9 N$

$F = 182.3 Lb$

Part b)

Work done to climb up to 3 m height is given by

$W = mgh$

now we have

$Power = \frac{Work}{time}$

$P = \frac{mgh}{t}$

$P = \frac{(70kg)(9.81)(3)}{5.0s}$

$P = 412.02 Watt$

now we know that 1 HP = 746 Watt

so we have

$P = \frac{412}{746} = 0.55 HP$

8 0

3 years ago

A 50.0-kg girl stands on a 9.0-kg wagon holding two 13.5-kg weights. She throws the weights horizontally off the back of the wag

Kobotan [32]

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5 0

3 years ago