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mrs_skeptik [129]

3 years ago

6

During a demonstration of the gravitational force on falling objects to her class, Sarah drops an 11 lb. bowling ball from the t

op of the science building. Determine distance the ball has traveled after falling for 1.0 second, ignoring air resistance and given the gravitational acceleration of 9.8 m/sec2.
A) 4.9 m
B) 11 m
C) 20 m
D) 32 m

The center on a high school basketball team wants to increase the maximum height of his jump. What statement correctly describes how he could achieve this goal?
A) Wearing lighter shoes would increase the height of his jump.
B) Jumping at a 45° angle would increase the height of his jump.
C) Jumping with greater acceleration would increase the height of his jump.
D) Jumping with greater initial velocity would increase the height of his jump

2 answers:

asambeis [7]3 years ago

7 0

1.

Answer: option A) 4.9 m

Explanation:

From the equation of motion,

s = u t + 0.5 a t ²

Where, t is the time, a is the acceleration, u is the initial velocity and s is the displacement.

Here, the initial velocity of the bowling ball is zero. u = 0

The ball falls under gravity. Thus, a = g = 9.8 m/s²

Time = 1.0 s

Hence, s = 0 + 0.5 × 9.8 m/s² × (1.0 s)² = 4.9 m

Thus, the ball has traveled 4.9 m after falling for 1.0 second. Correct option is A.

2.

Answer: option D) Jumping with greater initial velocity would increase the height of his jump

Explanation:

The maximum height achieved by a jumper depends on the angle , the initial velocity and acceleration due to gravity.

For maximum height, angle of the jump should be 90 degrees. Hence, more the initial velocity more the height of the jump.

Thus, correct option is D.

tia_tia [17]3 years ago

5 0

1.A) 4.9 m

AL2006 Ace

The instant it was dropped, the ball had zero speed.

After falling for 1 second, its speed was 9.8 m/s straight down (gravity).

Its AVERAGE speed for that 1 second was (1/2) (0 + 9.8) = 4.9 m/s.

Falling for 1 second at an average speed of 4.9 m/s, is covered 4.9 meters.

ANYTHING you drop does that, if air resistance doesn't hold it back.

Read more on Brainly.com - brainly.com/question/11776597#readmore

2 idk sorry

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4 years ago

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Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

5 0

3 years ago

Consider a 40,000 km steel pipe in the shape of a ring that fits snuggly all around the circumference of the Earth. We are heati

Tatiana [17]

Answer:

The Height is H = 70.02 m

Explanation:

We are given that the

Initial length is = $40000\ Km$ = $40,000 *10^{3} m$

from what we are told in the question the circumference of the circle is = $40,000 Km$

This means that the Radius would be :

Let C denote the circumference

So

$C = 2 \pi r$

=> $r = \frac{C}{2 \pi}$

$r = \frac{40,000}{2 \pi } = \frac{40,000*10^{3}}{2 *3.142} = 6.365*10^6 m$

We are told that 1-meter bar of steel that increases its temperature by 1 degree C will expand $11*10^{-6}$ meters

Hence

The final length would be

$40000*10^3 *(T + \alpha )$

Where T is the change in temperature $\alpha$ is the Coefficient of linear expansion for steel

let $L_{final}$ denote the final length

So

$L_{final} =40000*10^{6} *[1+ 11*10^{-6}]$

$= 40000440 \ m$

Now the Height is mathematically represented as

$Height(H) \ = \frac{change \ in \ radius \ }{2 \pi}$

$= \frac{(40000440-40000*10^3)}{2*3.142}$

$= 70.02m$

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