A 0.5 kg basketball moving 5 m/s to the right collides with a 0.05 kg tennis

A force of 85 N is used to push a box along the floor a distance of 15 m. How much work was done?

denpristay [2]

Answer:

1275J

Explanation:

Given parameters:

Force on box = 85N

Distance moved = 15m

Unknown:

Work done = ?

Solution:

Work done is the amount of force applied on a body to move it through a specific distance.

Work done = Force x distance

Now insert the parameters and solve;

Work done = 85 x 15 = 1275J

4 0

3 years ago

Read 2 more answers

Written:

zmey [24]

Answer:

0.833

Explanation:

Power = energy / time

Power = force × distance / time

Power = force × velocity

P = (850 kg) (9.8 m/s²) (1.00 m/s)

P = 8330 W

P = 8.33 kW

The efficiency of the motor is therefore:

e = 8.33 kW / 10.0 kW

e = 0.833

5 0

3 years ago

How does the end point differ from the equivalence point of a titration?

Gwar [14]

Answer:

Equivalence point and end point are terminologies in pH titrations and they are not the same. 

Explanation:

In a titration the substance added slowly to a solution usually through a pippette is called titrante and the solution to which it is added is called titrand. In acid-base titrations acid is added to base or base is added to acid.the strengths of the acid and base titrated determines the nature of the final solution.

At equivalence point the number of moles of the acid will be equal to the number of moles of the base as given in the equation. The nature of the final solution determines the pH at equivalence point. 

A pH less than 7 will be the result if the resultant is acidic and if it is basic the pH will be greater than 7. In a strong base-strong acid and weak base-weak acid titration the pH at the equivalence point will be 7 indicating neutral nature of the solution. 

3 0

3 years ago

NEED HELP!! ASAP JUST QUESTIONS 22, 23, 24, & 25

poizon [28]

22. reduction
25. Le Chatelier's principle

8 0

4 years ago

A car travels 40 miles in 30 minutes.

lukranit [14]

Answer:

(a)Average velocity ,v =128.74 Km/hr

(b)Kinetic Energy , K=958546.875 Joule

(c)Distance, s=268.8m

(d)Acceleration, a= - 2.38 $m/s^2$

Explanation:

Given:

Distance travelled = 40 miles

Time taken = 30 minutes.

(A) The average velocity in kilometres/hour

Converting 40 miles into km ,

we know that,

1 mile = 1.60934

40 miles = 40 x 1.60934

so 40 miles = 64.3738 Km

similarly converting 30 minutes into hours

1 minute = $\frac{1}{60}hours$

30 minute = $\frac{30}{60}hours$

30 minute = $\frac{1}{2}hours$

Now

Average velocity = $\frac{Speed}{time}$

Substituting Values,

Average velocity = $\frac{64.3738}{\frac[1}{2}}$

Average velocity = $64.3738 \times 2$

Average velocity =128.74 Km/hr

(B) If the car weighs 1.5 tons, what is its If the car weighs 1.5 tons, what is its kinetic energy in joules (Note: you will need to convert your velocity to m/s)? in joules (Note: you will need to convert your velocity to m/s)?

Converting 1.5 tons into kg we get

1 ton = 1000 kg

so 1.5 ton =1500 kg

converting velocity to m/s

$128.74 \times \frac{5}{18}$