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Gekata [30.6K]
4 years ago
12

A student of mass 40kg takes 10s to run up a flight of 50steps. If each step is 15cm high, calculate the Potential Energy of the

student at the maximum height.​
Physics
1 answer:
astra-53 [7]4 years ago
3 0

Answer:

2943 J

Explanation:

Potential energy = mass x acceleration due to gravity x height in meters

(P.E. = mgh)

Substitute the given numbers: (I take acceleration due to gravity as 9.81 m s^-2)

PE = 40 x 9.81 x (0.15x50)

PE = 2943 J

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which instrument is best suited for measuring the dimensions of a shoebox? a)a triple-beam balance b)a volumetric flask c)a rule
Aloiza [94]
I believe the correct answer from the choices listed above is option C. The instrument that is <span>best suited for measuring the dimensions of a shoebox would be a ruler. A triple-beam balance is for measuring mass. A volumetric flask is for volume. A caliper is measuring lengths of small objects.</span>
3 0
3 years ago
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A body weighs 100newtons when submerged in water. calculate the upthrust action on the body​
Andrews [41]

Answer:

Upthrust = 20 N

Explanation:

The question says that "A body weighs 100N in air and 80N when submerged in water. Calculate the upthrust acting on the body ?"

Upthrust is defined as the force when a body is submerged in liquid, then liquid applies a force on it.

ATQ,

Weight of body in air is 100 N

Weight of body in water is 80 N

Upthrust is equal to the weight of body in air minus weight of body in water.

Upthrust = 100 N - 80 N

Upthrust = 20 N

So, 20 N of upthrust is acting on the body.

7 0
3 years ago
You go to another forest and measure the shadow of a tree as being 6 meters long. The shadow of your meter
krok68 [10]
Well, 6/n=3/1, so n=2.
6 0
3 years ago
A 200.0 g block rests on a frictionless, horizontal surface. It is pressed against a horizontal spring with spring constant 4500
kenny6666 [7]

Answer:

 6 m/s

Explanation:

Given that :

mass of the block   m =  200.0 g  = 200 × 10⁻³ kg

the horizontal spring constant   k  =  4500.0 N/m

position of the block (distance x) = 4.00 cm  = 0.04 m

To determine the speed the block will be traveling when it leaves the spring; we applying the  work done on the spring as it is stretched (or compressed) with the kinetic energy.

i.e \frac{1}{2} kx^2  = \frac{1}{2} mv^2

kx^2 = mv^2

4500* 0.04^2 = 200*10^{-3} *v^2

7.2 =200*10^{-3}*v^{2}

v^{2}   =\frac{7.2}{200*10^{-3}}

v   =\sqrt{\frac{7.2}{200*10^{-3}}}

v = 6 m/s

Hence,the speed the block will be traveling when it leaves the spring is  6 m/s

5 0
4 years ago
Two long, parallel wires carry unequal currents in the same direction. The ratio of the currents is 3 to 1. The magnitude of the
astraxan [27]

Answer:

3A is the larger of the two currents.

Explanation:

Let the currents in the two wires be I₁ and I₂

given:

Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T

Distance, R = 10cm = 0.1m

Ratio of the current = I₁ : I₂ = 3 : 1

Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as

B = \frac{\mu_oI}{2\pi R}

Where \mu_o is the magnitude constant = 4π×10⁻⁷ H/m

Thus, the magnitude of a magnetic field due to I₁ will be

B_1 = \frac{\mu_oI_1}{2\pi R}

B_2 = \frac{\mu_oI_2}{2\pi R}

given,

B = B₁ - B₂ (since both the currents are in the same direction and parallel)

substituting the values of B, B₁ and B₂

we get

4.0×10⁻⁶T =  \frac{\mu_oI_1}{2\pi R} - \frac{\mu_oI_2}{2\pi R}

or

4.0×10⁻⁶T =  \frac{\mu_o}{2\pi R}\times (I_1-I_2 )

also

\frac{I_1}{I_2} = \frac{3}{1}

⇒I_1 = 3\times I_2

substituting the values in the above equation we get

4.0×10⁻⁶T =  \frac{4\pi\times 10^{-7}}{2\pi 0.1}\times (3 I_2-I_2)

⇒I_2 = 1A

also

I_1 = 3\times I_2

⇒I_1 = 3\times 1A

⇒I_1 = 3A

Hence, the larger of the two currents is 3A

3 0
3 years ago
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