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2 years ago

The force of friction on a sliding object is 10 n. the applied force needed to maintain a constant velocity is

1 answer:

8 0

Applied force should be equal to 10n so that there is no acceleration. When there is no acceleration, velocity is constant

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the minute hand on a clock is 9 cm long and travels through an arc of 252 degrees every 42 minutes. To the nearest tenth of a ce

Tanzania [10]

Answer:

The minutes hand travels 39.60 cm.

Explanation:

Note: a clock has a shape of a circle, the minutes hand is the radius, and the travel of the minutes hand forms a arc.

Length of an arc = ∅/360(2πr)

L = ∅/360(2πr).................... Equation 1π

Where L = length of an arc, ∅ = angle formed by an arc, r = radius of the arc.

Given: ∅ = 252°, r = 9 cm, π = 3.143.

Substituting these values into equation 1,

L = 252/360(2×3.143×9)

L = 0.7×2×3.143×9

L = 39.60 cm.

Thus the minutes hand travels 39.60 cm.

4 0

2 years ago

A basketball with mass of 0.8 kg is moving to the right with velocity 6 m/s and hits a volleyball with mass of 0.6 kg that stays

IceJOKER [234]

Answer:

26.67 m/s

Explanation:

From the law of conservation of linear momentum, the initial sum of momentum equals the final sum.

p=mv where p is momentum, m is the mass of object and v is the speed of the object

Initial momentum

The initial momentum will be that of basketball and volleyball, Since basketball is initially at rest, its initial velocity is zero

$p_i= m_bv_b+m_vv_v=8*6+0.6*0=48 Kg.m/s$

Final momentum

$p_f= m_bv_b+m_vv_v=8*4+0.6*v_v=32+0.6v Kg.m/s\\32+0.6v_v=48\\0.6v=16\\v_v=16/0.6=26.66666667\approx 26.67 m/s$

4 0

2 years ago

A 35 g steel ball is held by a ceiling-mounted electromagnet 4.0 m above the floor. A compressed-air cannon sits on the floor, 4

HACTEHA [7]

Answer:

7.9 m/s

Explanation:

When both balls collide, they have spent the same time for their motions.

Motion of steel ball

This is purely under gravity. It is vertical.

Initial velocity, u = 0 m/s

Distance, s = 4.0 m - 1.2 m = 2.8 m

Acceleration, a = g

Using the equation of motion

$s = ut+\frac{1}{2}at^2$

$2.8 \text{ m} = 0+\dfrac{gt^2}{2}$

$t = \sqrt{\dfrac{5.6}{g}}$

Motion of plastic ball

This has two components: a vertical and a horizontal.

The vertical motion is under gravity.

Considering the vertical motion,

Initial velocity, u = ?

Distance, s = 1.2 m

Acceleration, a = -g (It is going up)

Using the equation of motion

$s = ut+\frac{1}{2}at^2$

$1.2\text{ m} = ut-\frac{1}{2}gt^2$

Substituting the value of t from the previous equation,

$1.2\text{ m} = u\sqrt{\dfrac{5.6}{g}}-\dfrac{1}{2}\times g\times\dfrac{5.6}{g}$

$u\sqrt{\dfrac{5.6}{g}} = 4.0$

Taking g = 9.8 m/s²,

$u = \dfrac{4.0}{0.756} = 5.29 \text{ m/s}$

This is the vertical component of the initial velocity

Considering the horizontal motion which is not accelerated,

horizontal component of the initial velocity is horizontal distance ÷ time.

$u_h = \dfrac{4.4\text{ m}}{0.756\text{ s}} = 5.82\text{ m/s}$

The initial velocity is

$v_i = \sqrt{u^2+u_h^2} = \sqrt{(5.29\text{ m/s})^2+(5.82\text{ m/s})^2} = 7.9 \text{ m/s}$

4 0

2 years ago

A diver explores a shallow reef off the coast of Belize. She initially swims d1 = 74.8 m north, makes a turn to the east and con

Nataly [62]

Answer:R=1607556m

θ=180degrees

Explanation:

d1=74.8m

d2=160.7km=160.7km*1000

d2=160700m

d3=80m

d4=198.1m

Using analytical method :

Rx=-(160700+75*cos(41.8))= -160755.9m

Ry= -(74.8+75sin(41.8))-198.1=73m

Magnitude, R:

R=√Rx+Ry

R=√160755.9^2+20^2=160755.916

R=160756m

Direction,θ:

θ=arctan(Rx/Ry)

θ=arctan(-73/160755.9)

θ=-7.9256*10^-6

Note that θ is in the second quadrant, so add 180

θ=180-7.9256*10^6=180degrees

8 0

2 years ago

1. If an object on a horizontal frictionless surface is attached to a spring, displaced, and then released. It will oscillate. I

AURORKA [14]

Answer: a) 0.12m; b) 1,6 s; c) 0.625 1/s

Explanation: The simple harmonic movement can be described by a sin or cosine function in time.

This can be in the form:

X(t)= A Sin/Cos (wt+φ) where φ is initial phase o position at t=0

w the angular frequency are related to the frequency (f) as 2Pif

and f=1/T period of oscillating

3 0

2 years ago