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andrew11 [14]
3 years ago
11

Answer and I will give you brainiliest Please I need a surely answer ​

Physics
2 answers:
hichkok12 [17]3 years ago
8 0

Answer:

<h2><u>Constant</u></h2>

Explanation:

Please don't comment in this question's comment box

<h2>Thanks</h2>
tatyana61 [14]3 years ago
3 0

Answer:

constant

I hope im right !!!

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a student lift a 25kg mass at vertical distance of 1.6m in a time of 2.0 seconds. a. Find the force needed to lift the mass (in
zhannawk [14.2K]

Answer:

a. F = 245 Newton.

b. Workdone = 392 Joules.

c. Power = 196 Watts

Explanation:

Given the following data;

Mass = 25kg

Distance = 1.6m

Time = 2secs

a. To find the force needed to lift the mass (in N );

Force = mass * acceleration

We know that acceleration due to gravity is equal to 9.8

F = 25*9.8

F = 245N

b. To find the work done by the student (in J);

Workdone = force * distance

Workdone = 245 * 1.6

Workdone = 392 Joules.

c. To find the power exerted by the student (in W);

Power = workdone/time

Power = 392/2

Power = 196 Watts.

5 0
3 years ago
What mechanical layer lies below the lithosphere
joja [24]
The <span>asthenosphere is under the lithosphere.</span>
3 0
3 years ago
Read 2 more answers
For a spring oscillating in simple harmonic motion, at what point will the velocity of
Crazy boy [7]

Answer:

equilibrium position.

Explanation:

In simple harmonic motion , velocity v(t) is given by,

v(t) = -ω A sin(ωt + φ)

where

ω = angular velocity of the corresponding circular motion

A = amplitude

t = time

φ = the initial angle of the corresponding circular motion when the motion begin.

v (t) get maximized when sin value is maximized , i.e. sin \alpha=1

The particle has maximum speed when it passes through the equilibrium position.

6 0
3 years ago
A professional racecar driver buys a car that can accelerate at 5.9 m/s2. The racer decides to race against another driver in a
3241004551 [841]

Answer:

(a) Time will be t = 3.56 sec

(b) Distance traveled by car when they are side by side is 37.38712 m

(b) Velocity of race car = 21.004 m/sec

velocity of stock car = 12.816 m/sec            

Explanation:

We have given acceleration of the car a_1=5.9m/sec^2

Acceleration of the stock car a_2=3.6m/sec^2

When 1st car overtakes the second car then distance traveled by both the car will be same

(a) So s_1=s_2

As both car starts from rest so initial velocity of both car will be 0 m/sec

It is given that stock car leaves 1 sec before

So \frac{1}{2}\times 5.9\times t^2=\frac{1}{2}\times (t+1)^2\times 3.6

After solving t = 3.56 sec

(b) From second equation of motion s=ut+\frac{1}{2}at^2=0\times 3.56+\frac{1}{2}\times 5.9\times 3.56^2=37.38712m

(c) From first equation pf motion v = u+at

So velocity of race car v = 0+5.9×3.56 = 21.004 m/sec

Velocity of stock car v = 0+ 3.6×3.56 = 12.816 m/sec

3 0
3 years ago
The average Intensity of the electromagnetic field due to sunlight at the surface of the earth is 1400 W/m^2. Calculate the maxi
Ann [662]

Answer:

1027 N/C

3.42 x 10⁻⁶ T

Explanation:

I = Intensity of electromagnetic field = 1400 W/m²

E₀ = Maximum value of electric field

Intensity of electromagnetic field is given as

I = (0.5) ε₀ E₀² c

1400 =  (0.5) (8.85 x 10⁻¹²) (3 x 10⁸) E₀²

E₀ = 1027 N/C

B₀ = maximum value of magnetic field

using the equation

E₀ = B₀ c

1027 = B₀ (3 x 10⁸)

B₀ = 3.42 x 10⁻⁶ T

4 0
3 years ago
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