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3 years ago

10

What’s the difference between applied and pure research

1 answer:

Ket [755]3 years ago

8 0

Answer:

Pure research is conducted without any specific goal in mind. Applied research is conducted in order to solve a specific and practical problem, unlike pure research. Hope this helped.

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A 100 Kg man is diving off a 50 meter cliff. What is his kinetic energy when he is 20 meters from the water?

iren2701 [21]

Answer:

$K.E=29.403125J$

Explanation:

From the question we are told that

Mass $M=100$

Height $50-20=30m$

Generally the equation for velocity before impact is is is mathematically given by

$v=\sqrt{2gh}$

$v=\sqrt{2*9.8*30}$

$v=24.25$

Generally the equation for Kinetic Energy is is mathematically given by

$K.E=\frac{1}{2}mv^2$

$K.E=\frac{1}{2}*100*(24.25)^2\\$

$K.E=29403.125J$

$K.E=29.403125J$

8 0

3 years ago

A 12.8-kg monkey hangs from a cord suspended from the ceiling of an elevator. The cord can withstand a tension of 156N and break

Varvara68 [4.7K]

Answer:

$2.4 m/s^2$

Downward

Explanation:

We are given that

Mass of monkey=12.8 kg

Tension=156 N

We have to find the magnitude of the elevator's minimum acceleration.

$T=m(g+a)$

Where $g=9.8 m/s^2$

Substitute the values

$156=12.8(a+9.8)$

$9.8+a=\frac{156}{12.8}=12.2$

$a=12.2-9.8=2.4 m/s^2$

Hence, the acceleration = $a=2.4 m/s^2$

Direction of the elevator's minimum acceleration is downward because the elevator moves downwards.

8 0

3 years ago

A certain car engine delivers enough force to create 630 N⋅m of torque when the engine is operating at 3200 revolutions per minu

jekas [21]

The appropriate expression for the calculation of power by relating the angular energy in a given time.

In other words the instantaneous power of an angular accelerating body is the torque times the angular velocity

$P=\tau\omega$

Where

$\tau =$ Torque

$\omega =$ Angular speed

Our values are given by

$\tau = 630Nm$

$\omega = 3200rev/min$

The angular velocity must be transformed into radians per second then

$\omega = 3200rev/min (\frac{2\pi rad}{60s})$

$\omega = 335.103rad/s$

Replacing,

$P=(630)(335.103)$

$P = 211.11*10^3W$

$P = 211.1kW$

The average power delivered by the engine at this rotation rate is 211.1kW

5 0

3 years ago

car 2 has a mass of 150 kg and moves westward towards car 3 at a velocity of 2.2 m/s. car 3 has a mass of 265 kg and moves eastw

sergejj [24]

Answer:

The force of car 3 on car 2 ≈ 1810.82 N

Explanation:

The equation for the change in momentum of the two cars are;

Conservation of linear momentum

150( 2.2 - v) = 265(1.5-v)

150 × 2.2 - 265×1.5 = (150+265)v

150 × 2.2 - 265×1.5 = -67.5 = 415×v

∴ v = -67.5/415 = -0.1627 m/s West = 0.1627 m/s East

The impulse of the net force is the amount of momentum change experienced given by the equation;

Impulse force = $m \times v_f$ - $m \times v_0$

Where;

$v_f$ = The final velocity

$v_0$ = The initial velocity

For the the 265 kg mass, we have;

$v_f$ = 0.1627 m/s

$v_0$ = 1.5 m/s

Which gives the impulse a s F×Δt = 265×0.1627 - 265×1.5 = -354.38 kg·m/s

The change in kinetic energy of the collision = 1/2×265×(0.1627^2 - 1.5^2) =-294.62 J

Whereby the distance moved in one second is 0.1627 m, we have;

Work done = Force × Distance = Force × 0.1627 = 294.62

Force = 294.62/0.1627 = 1810.82 N.

8 0

3 years ago

On Mars, where air resistance is negligible, an astronaut drops a rock from a cliff and notes that the rock falls about d meters

dimulka [17.4K]

Answer:

$d_1 = 16 d$

Explanation:

As we know that initial speed of the fall of the stone is ZERO

$v_i = 0$

also the acceleration due to gravity on Mars is g

so we have

$d = v_i t + \frac{1}{2}gt^2$

now we have

$d = 0 + \frac{1}{2}g t^2$

now if the same is dropped for 4t seconds of time

then again we will use above equation

$d_1 = 0 + \frac{1}{2}g(4t)^2$

$d_1 = 16(\frac{1}{2}gt^2)$

$d_1 = 16 d$

7 0

3 years ago

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