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Karo-lina-s [1.5K]

3 years ago

14

A skateboarder has an average velocity of 5.0 m/s to the right. Does this mean that the skateboarder moves with a constant veloc

ity of + 5.0 m/s? Explain.

1 answer:

Ainat [17]3 years ago

8 0

No this doesn't mean constant velocity of 5m/s. as his speeds could be a little over 5 and under 5 to be an average of 5. so you can't be sure, so no

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What's √ 25what's √ 25

malfutka [58]

Answer:

the square root of 25 is 5

Explanation:

$\sqrt{25}=5$

5 0

3 years ago

Radioactive iodine (radioiodine) is given to patients with some forms of thyroid cancer in order to destroy the cancer cells. 13

Eva8 [605]

Answer:

I think is D

Explanation:

Please mark me as brainliest

4 0

3 years ago

In April 1974, Steve Prefontaine completed a 10.0 km race in a time of 27 min , 43.6 s . Suppose "Pre" was at the 7.43 km mark a

Novay_Z [31]

Answer:

Acceleration, a = 0.101 m/s²

Explanation:

Average speed = total distance / total time.

At the 7.43km mark, total distance = 7.43km or 7430m

Total time = 25 * 60 s = 1500s

Average speed = 7430m/1500s = 4.95m/s

He then covers (10 - 7.43)km = 2.57 km = 2570 m

in t = 27m43.6s - 25min = 2m43.6s = 163.6 s

Then he accelerates for 60 s, and maintains this velocity V, for the remaining (163.6 - 60)s = 103.6 s.

From V = u + at; V = 4.95m/s + a *60s

Distance covered while accelerating is

s = ut + ½at² = 4.95m/s * 60s + ½ a *(60s)² = 297m + a*1800s²

Distance covered while at constant velocity, v after accelerating is

D = velocity * time

Where v = 4.95m/s + a*60s

D = (4.95m/s + a*60s) * 103.6s = 512.82m + a*6216s²

Total distance covered after initial 7.43 km, S + D = 2570 m, so

2570 m = 297m + a*1800s² + 512.82m + a*6216s²

2570 = 809.82 + a*8016

a = 809.82m / 8016s² = 0.101 m/s²

8 0

3 years ago

- A cannon of 2000 kg fires a shell of 10 kg at

fgiga [73]

1) -0.5 m/s

We can solve the first part of the problem by using the law of conservation of momentum. In fact, the total momentum of the cannon - shell system must be conserved.

Before the shot, both the cannon and the shell are at rest, so the total momentum is zero:

$p=0$

After the shot, the momentum is:

$p=MV+mv$

where

M = 2000 kg is the mass of the cannon

m = 10 kg is the mass of the shell

v = 100 m/s is the velocity of the shell (we take as positive the direction of motion of the shell)

V = ? is the velocity of the cannon

Since momentum is conserved, we can write

$0=MV+mv$

And solving for V, we find the velocity of the cannon:

$V=-\frac{mv}{M}=-\frac{(10)(100)}{2000}=-0.5 m/s$

where the negative sign indicates that the cannon moves in the direction opposite to the shell.

2) 0.5 m

The motion of the cannon is a uniformly accelerated motion, so we can solve this part by using suvat equation:

$v^2-u^2=2as$

where

v is the final velocity of the cannon

u = 0.5 m/s is the initial velocity of the cannon (now we take as positive the initial direction of motion of the cannon)

$a=-0.25 m/s^2$ is the deceleration of the cannon

s is the distance travelled by the cannon

The cannon will stop when v = 0; substituting and solving the equation for s, we find the minimum safe distance required to stop the cannon:

$s=\frac{v^2-u^2}{2a}=\frac{0-0.5^2}{2(-0.25)}=0.5 m$

7 0

4 years ago

What is the outcome of the Training and Exercise Planning Workshop (TEPW)?

balandron [24]

Answer / Explanation:

The result of the Training and Exercise Planning Workshop (TEPW) is to set the foundation for the strategy and pattern for a proposed exercise program. The TEPW purpose is to engage elected and selected officials in identifying exercise program priorities and planning a schedule of training and exercise events to meet those priorities.

An essential factor for the exercise management process is to create a collaborative environment where a whole community stakeholders can engage in a forum to discuss and coordinate training and exercise activities across local organizations to maximize the use of available resources and prevent duplication of effort.

8 0

3 years ago