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3 years ago

A bus is moving forward at 20 m/s. A student on the bus throws a tenis ball horizontally at

Physics

1 answer:

seraphim [82]3 years ago

8 0

Answer:

D: 35 m/s

Explanation:

The bus is moving at a speed of 20 m/s.

Thus; v_bus = 20 m/s

Tennis ball thrown horizontally towards the front of the bus is given as 15 m/s.. Thus, v_ball = 15 m/s

No, due to the fact that the bus and the ball are moving at the same time, an observer will think the speed is the sum of that of the ball and the bus.

Thus, it will appear to an observer on the sidewalk that the speed is; v_bus + v_ball = 20 + 15 = 35 m/s

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And instead of tolling the bell, for church, our little sexton – sings. what is the most likely reason for the poet to oppose th

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The poet is expressing her belief that the second, more natural option is far more desirable than the first option.

What is the summary of some keep the sabbath going to church?

Some hold the Sabbath going to Church is set faith and meditation.

The poem explores the idea of spiritual practices in Christianity.

It additionally illustrates the speaker's angle in the direction of those practices.

According to this, it's far obligatory to put on a surplice even as attending church services.

While she wears her casual get dressed and prays to God, Some hold the Sabbath going to Church –' is one in every of Emily Dickinson's best-recognized poems.

It functions her aim to are seeking for salvation without resorting to the traditional means.

hence,

The most likely reason for the poet to oppose the phrases "tolling the bell" and "sings" in these lines is the poet is expressing her belief that the second, more natural option is far more desirable than the first option.

Thus, option "A" is correct.

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2 years ago

A candle is 49 cm in front of a convex spherical mirror that has a focal length

vitfil [10]

The convex spherical mirror is the diverging mirror. The image distance is 20.4 cm. The magnification is 0.417. The image is virtual and upright.

<h3>What is magnification?</h3>

Magnification is the ratio of image distance to the object distance.

Object distance is 49 cm and the focal length of convex mirror is 35 cm, then image distance is calculated by the lens maker formula.

1/f = 1/v +1/u

1/-35 = 1`/v +1/49

v= -20.4 cm

Thus, the image distance is 20.4 cm.

The magnification is given by

m = v/u

m = -20.4 / 49

m = 0.417

Thus, the magnification is 0.417.

The image formed is by virtual meeting of light rays and above the principle axis. Thus the image is virtual and upright.

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3 years ago

A sphere of mass m and radius r is released from rest at the top of a curved track of height H. The sphere travels down the curv

iren2701 [21]

Explanation:

(a) On the dots below, which represent the sphere, draw and label the forces (not components) that are exerted on the sphere at point A and at point B, respectively. Each force must be represented by a distinct arrow starting on and pointing away from the dot.

At point A, there are three forces acting on the sphere: weight force mg pulling down, normal force N pushing left, and static friction force Fs pushing down.

At point B, there are three forces acting on the sphere: weight force mg pulling down, normal force N pushing down, and static friction force Fs pushing right.

(b) i. Derive an expression for the speed of the sphere at point A.

Energy is conserved:

PE = PE + KE + RE

mgH = mgR + ½mv² + ½Iω²

mgH = mgR + ½mv² + ½(⅖mr²)(v/r)²

mgH = mgR + ½mv² + ⅕mv²

gH = gR + ⁷/₁₀ v²

v² = 10g(H−R)/7

v = √(10g(H−R)/7)

ii. Derive an expression for the normal force the track exerts on the sphere at point A.

Sum of forces in the radial (-x) direction:

∑F = ma

N = mv²/R

N = m (10g(H−R)/7) / R

N = 10mg(H−R)/(7R)

(c) Calculate the ratio of the rotational kinetic energy to the translational kinetic energy of the sphere at point A.

RE / KE

= (½Iω²) / (½mv²)

= ½(⅖mr²)(v/r)² / (½mv²)

= (⅕mv²) / (½mv²)

= ⅕ / ½

= ⅖

(d) The minimum release height necessary for the sphere to travel around the loop and not lose contact with the loop at point B is Hmin. The sphere is replaced with a hoop of the same mass and radius. Will the value of Hmin increase, decrease, or stay the same? Justify your answer.

When the sphere or hoop just begins to lose contact with the loop at point B, the normal force is 0. Sum of forces in the radial (-y) direction:

∑F = ma

mg = mv²/R

gR = v²

Applying conservation of energy:

PE = PE + KE + RE

mgH = mg(2R) + ½mv² + ½Iω²

mgH = 2mgR + ½mv² + ½(kmr²)(v/r)²

mgH = 2mgR + ½mv² + ½kmv²

gH = 2gR + ½v² + ½kv²

gH = 2gR + ½v² (1 + k)

Substituting for v²:

gH = 2gR + ½(gR) (1 + k)

H = 2R + ½R (1 + k)

H = ½R (4 + 1 + k)

H = ½R (5 + k)

For a sphere, k = 2/5. For a hoop, k = 1. As k increases, H increases.

(e) The sphere is again released from a known height H and eventually leaves the track at point C, which is a height R above the bottom of the loop, as shown in the figure above. The track makes an angle of θ above the horizontal at point C. Express your answer in part (e) in terms of m, r, H, R, θ, and physical constants, as appropriate. Calculate the maximum height above the bottom of the loop that the sphere will reach.

C is at the same height as A, so we can use our answer from part (b) to write an equation for the initial velocity at C.

v₀ = √(10g(H−R)/7)

The vertical component of this initial velocity is v₀ sin θ. At the maximum height, the vertical velocity is 0. During this time, the sphere is in free fall. The maximum height reached is therefore:

v² = v₀² + 2aΔx

0² = (√(10g(H−R)/7) sin θ)² + 2(-g)(h − R)

0 = 10g(H−R)/7 sin²θ − 2g(h − R)

2g(h − R) = 10g(H−R)/7 sin²θ

h − R = 5(H−R)/7 sin²θ

h = R + ⁵/₇(H−R)sin²θ