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2 years ago

A body accelerates by 25m/s 2 when it applied by 40n forces.what would be acceleration if it is applied by 80n forces

Physics

2 answers:

frutty [35]2 years ago

7 0

Answer:

50m/s2

Explanation:

since the force is doubled the acceleration will be doubled

il63 [147K]2 years ago

6 0

Answer:

50m/s²

Explanation:

cross muliplyy write down the values acceleration and force

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What is the magnitude of your displacement when you follow directions that tell you to walk 100.0m north, then 25.0m East?

atroni [7]

Answer:

Displacement from the starting position is 103.21m

Explanation:

If you draw these directions, it will create the two legs of a triangle.

Using this method, you can visualize why your displacement is what it is.

Using the pythagorean theorem

${a}^{2} + {b}^{2} = {c}^{2}$

Plug in both values

${100m}^{2} + {25m}^{2} = {c}^{2}$

$c = \sqrt{10652}$

c = 103.2085

c= 103.21

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3 years ago

How does beam spreading affect the temperature on earth and the seasons

navik [9.2K]

Get to school and learn boi

8 0

2 years ago

A 1.50-V battery supplies 0.414 W of power to a small flashlight. If the battery moves 4.93 1020 electrons between its terminals

slamgirl [31]

Answer:

2.86×10⁻¹⁸ seconds

Explanation:

Applying,

P = VI................ Equation 1

Where P = Power, V = Voltage, I = Current.

make I the subject of the equation

I = P/V................ Equation 2

From the question,

Given: P = 0.414 W, V = 1.50 V

Substitute into equation 2

I = 0.414/1.50

I = 0.276 A

Also,

Q = It............... Equation 3

Where Q = amount of charge, t = time

make t the subject of the equation

t = Q/I.................. Equation 4

From the question,

4.931020 electrons has a charge of (4.931020×1.6020×10⁻¹⁹) coulombs

Q = 7.899×10⁻¹⁹ C

Substitute these value into equation 4

t = 7.899×10⁻¹⁹/0.276

t = 2.86×10⁻¹⁸ seconds

5 0

2 years ago

Find the velocity of the student using KE= 1/2 mv^2. A 50-kilogram student is running and has 225 joules of kinetic energy. The

defon

225 = 1/2 (50) (v2)
225 = 25 (v2)
225/25 = v2
9 = v2
√9 = v
v = 3 m/s

3 0

3 years ago

Someone help please.....

Westkost [7]

Answer:

0.0928km/min (4dp)

Explanation:

To find the jogger's speed in km per minute, we just need to divide the number of km jogged by the time in minutes it took to jog that distance. This will give us the distance they jogged every minute which is their speed.

4km in 32 minutes:

4/32 = 0.125km/min

2km in 22 minutes:

2/22 = 0.091 (3dp)km/min

1km in 16 minutes:

0.0625km/min

Now to find the average speed of these 3 speeds, we just add them all together and divide by how many values there are (3 values).

Average (mean) = $\frac{0.125+0.091+0.0625}{3}$

Average = 0.2785/3

Average speed of jogger = 0.0928 (4dp) km/min

Hope this helped!

8 0

3 years ago