SOMEONE HELP ME PLEASE

A soccer player is running upfeild at 10m/s and comes to a stop in 3 seconds facing in the same direction what is his accelerati

kumpel [21]

-3.33 m/s^2

The soccer player decelerates from 10 m/s to 0 m/s. Because he decelerates the acceleration will be a negative value. Since he takes 3 seconds to decelerate, divide his original velocity by 3

One major point to be careful of is to keep p

6 0

4 years ago

An engine that has an efficiency of 25% takes in 200 [J] of heat during each cycle. Calculate the amount of work this engine per

BartSMP [9]

Answer:

$W=50J$

Explanation:

The work done $W$ and the energy taken $Q_{in}$ by a heat engine are related to the efficiency $\eta$ by the expression

$\eta=\frac{W}{Q_{in}}$

The efficiency is $\eta$ =25%, the numerical form of this percentage is 0.25 and the energy taken is $Q_{in}$ =200J. Replacing in the formula:

$0.25=\frac{W}{200}\\W=200*0.25\\W=50J$

5 0

3 years ago

What best describes the relationship between wavelength and frequency in an electromagnetic wave?

netineya [11]

Inversely proportional
I hope this helps you!

4 0

4 years ago

Read 2 more answers

Stars give off a tremendous amount of thermal and ________ energy.(1 point) chemical chemical electrical electrical nuclear nucl

Olegator [25]

I believe the answer is Nuclear energy

8 0

2 years ago

A dockworker applies a constant horizontal force of 80.0 N to a block of ice on a smooth horizontal floor. The frictional force

Tamiku [17]

Answer:

(a) 91 kg (2 s.f.) (b) 22 m

Explanation:

Since it is stated that a constant horizontal force is applied to the block of ice, we know that the block of ice travels with a constant acceleration and but not with a constant velocity.

(a)

$s \ = \ ut \ + \ \displaystyle\frac{1}{2} at^{2} \\ \\ a \ = \ \displaystyle\frac{2(s \ - \ ut)}{t^{2}} \\ \\ a \ = \ \displaystyle\frac{2(11 \ - \ 0)}{5^{2}} \\ \\ a \ = \ \displaystyle\frac{22}{25} \\ \\ a \ = \ 0.88 \ \mathrm{m \ s^{-2}}$

Subsequently,

$F \ = \ ma \\ \\ m \ = \ \displaystyle\frac{F}{a} \\ \\ m \ = \ \displaystyle\frac{80 \ \mathrm{kg \ m \ s^{-2}}}{0.88 \ \mathrm{m \ s^{-2}}} \\ \\ m \ = \ 91 \mathrm{kg} \ \ \ \ \ \ (2 \ \mathrm{s.f.})$

*Note that the equations used above assume constant acceleration is being applied to the system. However, in the case of non-uniform motion, these equations will no longer be valid and in turn, calculus will be used to analyze such motions.

(b) To find the final velocity of the ice block at the end of the first 5 seconds,

$v \ = \ u \ + \ at \\ \\ v \ = \ 0 \ + \ (0.88 \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \\ \\ v \ = \ 4.4 \ \mathrm{m \ s^{-1}}$

According to Newton's First Law which states objects will remain at rest

or in uniform motion (moving at constant velocity) unless acted upon by

an external force. Hence, the block of ice by the end of the first 5

seconds, experiences no acceleration (a = 0) but travels with a constant

velocity of 4.4 $m \ s^{-1}$ .

$s \ = \ ut \ + \ \displaystyle\frac{1}{2}at^{2} \\ \\ s \ = \ (4.4 \ \mathrm{m \ s^{-2}})(5 \ \mathrm{s}) \ + \ \displaystyle\frac{1}{2}(0)(5^{2}) \\ \\ s \ = \ 22 \ \mathrm{m}$

Therefore, the ice block traveled 22 m in the next 5 seconds after the

worker stops pushing it.

4 0

3 years ago