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mylen [45]
2 years ago
11

How much force is needed to accelerate a box of mass 25kg at 4 m/s2

Physics
1 answer:
olga_2 [115]2 years ago
5 0

Answer:

D

Explanation:

Force (N) = mass (kg) × acceleration (m/s²) is the famous F=ma

Force = 25 * 4 in this case

Force = 100 N

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HELP MEEEEEEE!<br><br> Compare fossils x and y during tension. What happens to each?
Kruka [31]

Answer:

The layers will shift

Explanation:

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3 years ago
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What is the mass of 5 moles of gold
Serhud [2]
First of all, we know that one mole is equal to the atomic number of an element.

The atomic number of gold is <span>197.0g Au

And we need to find 5 moles.

5 * 197.0 g Au = </span><span>985.0g

Grams is used to measure mass. 

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3 years ago
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On a hot summer day, 3.50 ✕ 106 J of heat transfer into a parked car takes place, increasing its temperature from 36.5°C to 44.4
anygoal [31]

Answer:

a) \Delta s=443037.9747\ J.K^{-1}

b) \Delta s=31868131.8681\ J.K^{-1}

Explanation:

a)

Given:

amount of heat transfer occurred,dQ=3.5\times 10^6\ J

initial temperature of car, T_i=36.5+273=309.5\ K

final temperature of car, T_f=44.4+273=317.4\ K

We know that the change in entropy is given by:

\Delta s=\frac{dQ}{T_f-T_i}

\Delta s=\frac{3.5\times 10^6}{(44.4-36.5)} (heat is transferred into the system of car)

\Delta s=443037.9747\ J.K^{-1}

b)

amount of heat transfer form the system of house, dQ=5.8\times 10^8\ J

initial temperature of house, T_i=23.5+273=296.5\ K

final temperature of house, T_f=5.3+273=278.3\ K

\Delta s=\frac{dQ}{T_f-T_i}

\Delta s=\frac{5.8\times 10^8}{278.3-296.5}

\Delta s=31868131.8681\ J.K^{-1}

6 0
3 years ago
Which of the below is an example of mimicry that enables prey species avoid predation?
Marta_Voda [28]

b..a harmless organism imitating the look of a harmful organism

Explanation:

A harmless organism imitating the look of a harmful organism is one example of mimicry that enables prey species avoid predation.

Prey are smaller and less harmful organisms often hunted by larger organisms usually carnivores.

Mimicry is a form of evolutionary adaptation process in which two organisms of the same specie or different species tends to look alike.

It is a subtle defense mechanism developed by organism over an extended period of time.

learn more:

Adaptation brainly.com/question/11105547

#learnwithBrainly

6 0
3 years ago
A train travels due north in a straight line with a constant speed of 100 m/s. Another train leaves a station 2,881 m away trave
damaskus [11]

Answer:

The trains will collide at a distance 1660 m from the station

Explanation:

Let the train traveling due north with a constant speed of 100 m/s be Train A.

Let the train traveling due south with a constant speed of 136 m/s be Train B.

From the question, Train B leaves a station 2,881 m away (that is 2,881 m away from Train A position).

Hence, the two trains would have traveled a total distance of 2,881 m by the time they collide.

∴ If train A has covered a distance x m by the time of collision, then train B would have traveled (2881 - x) m.

Also,

At the position where the trains will collide, the two trains must have traveled for equal time, t.

That is, At the point of collision,

t_{A} = t_{B}

t_{A} is the time spent by train A

t_{B} is the time spent by train B

From,

Velocity = \frac{Distance }{Time }\\

Time = \frac{Distance}{Velocity}

Since the time spent by the two trains is equal,

Then,

\frac{Distance_{A} }{Velocity_{A} }  = \frac{Distance_{B} }{Velocity_{B} }

{Distance_{A} = x m

{Distance_{B} = 2881 - x m

{Velocity_{A} = 100 m/s

{Velocity_{B} = 136 m/s

Hence,

\frac{x}{100} = \frac{2881 - x}{136}

136(x) = 100(2881 - x)\\136x = 288100 - 100x\\136x + 100x = 288100\\236x = 288100\\x = \frac{288100}{236} \\x = 1220.76m\\

x≅ 1,221 m

This is the distance covered by train A by the time of collision.

Hence, Train B would have covered (2881 - 1221)m = 1660 m

Train B would have covered 1660 m by the time of collision

Since it is train B that leaves a station,

∴ The trains will collide at a distance 1660 m from the station.

7 0
3 years ago
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