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2 years ago

A 30kg boxed is pushed with a force of 20N. What is the boxes acceleration. Please show work

Physics

1 answer:

kozerog [31]2 years ago

8 0

Answer:

<h3>The answer is 0.67 m/s²</h3>

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question we have

$a = \frac{20}{30} = \frac{2}{3} \\ = 0.6666666...$

We have the final answer as

Hope this helps you

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What does law and theory have common

Ronch [10]

Answer:

It is a misconception that theories turn into laws with enough research.

Hope it helps

7 0

2 years ago

A hot-water bottle contains 787 g of water at 75∘C. If the liquid water cools to body temperature (37 ∘C), how many kilojoules o

IgorC [24]

Answer:

$Q = 787 gr * 1 \frac{cal}{gr C} *(37-75)C = -29906 cal$

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

$29906 cal *\frac{4.184 J}{1 cal}* \frac{1KJ}{1000 J}= 125.127 KJ$

Explanation:

For this case we know the mass of the water given :

$m = 787 gr$

And we know that the initial temperature for this water is $T_i =75 C$ .

We want to cool this water to the human body temperature $T_f = 37 C$

Since the temperatures given are not near to 0C (fusion point) or 100C (the boling point) we don't need to use latent heat, then the only heat involved for this case is the sensible heat given by:

$Q= m c_p \Delta T$

Where $c_p$ represent the specific heat for the water and this value from tables we know that $c_p =1 \frac{cal}{gr C}$ for the water.

So then we have everything in order to replace into the formula of sensible heat and we got:

$Q = 787 gr * 1 \frac{cal}{gr C} *(37-75)C = -29906 cal$

So then the answer for this case would be 29906 cal but we need to convert this into KJ and we know that 1 cal = 4.184 J and if we convert we got:

$29906 cal *\frac{4.184 J}{1 cal}* \frac{1KJ}{1000 J}= 125.127 KJ$

8 0

2 years ago

Read 2 more answers

You are driving home from school steadily at for 180 km. It then begins to rain and you slow to You arrive home after driving 4.

nata0808 [166]

Answer:

Explanation:

Question is incomplete

Assuming the question you have asked is

You are driving home from school steadily at 95 km/h for 180 km. It then begins to rain and you slow to 65 km/h. You arrive home after driving 4.5 h.

given,

speed of 95 km/h for 180 km

due to rain

speed is reduced to 65 km/h

distance traveled in 4.5 hour

time taken to travel 180 km

d = s x t

$t = \dfrac{180}{95}$

t = 1.9 hr

distance traveled in time, t' = 4.5-1.9 = 2.6 hr

Speed of vehicle = 65 Km/h

d' = s x t'

d' = 65 x 2.6

d'= 169 Km

total distance your hometown from school

D = d + d'

D = 180 + 169

D = 349 Km

6 0

3 years ago

What is the best way to dry a thermometer servsafe?

Mandarinka [93]

A thermometer should be air dried so as to not damage it or anything similar. The three steps to cleaning it are washing, rinsing, and air drying it after. You shouldn't try to dry it other ways because it can damage it and this can cause a lot of troubles, so things cold air blowers or similar things can be very good in cleaning your thermometer.

6 0

3 years ago

Water runs into a fountain, filling all the pipes, at a steady rate of 0.750 m3>s. (a) How fast will it shoot out of a hole 4

kati45 [8]

Answer:

velocity = 472 m/s

velocity = 52.4 m/s

Explanation:

given data

steady rate = 0.750 m³/s

diameter = 4.50 cm

solution

we use here flow rate formula that is

flow rate = Area × velocity .............1

0.750 = $\frac{\pi }{4}$ × (4.50× $10^{-2}$ )² × velocity

solve it we get

velocity = 472 m/s

and

when it 3 time diameter

put valuer in equation 1

0.750 = $\frac{\pi }{4}$ × 3 × (4.50× $10^{-2}$ )² × velocity

velocity = 52.4 m/s

5 0

2 years ago