1 st QUESTION: In the raisin experiment what was your first step of the scientific method?

Two particles, each of mass 7.0 kg, are a distance 3.0 m apart. To bring a third particle, with mass 21 kg, from far away to a r

garri49 [273]

Answer: the external agent must do work equal to -1.3 × 10⁻⁸ J

Explanation:

Given that;

Mass M1 = 7.0 kg

r = 3.0/2 m = 1.5 m

Mass M2 = 21 kg

we know that G = 6.67 × 10⁻¹¹ N.m²/kg²

work done by an external agent W = -2GM2M1 / r

so we substitute

W = (-2 × 6.67 × 10⁻¹¹ × 21 × 7) / 1.5

W = -1.96098 × 10⁻⁸ / 1.5

W = -1.3 × 10⁻⁸ J

Therefore the external agent must do work equal to -1.3 × 10⁻⁸ J

8 0

3 years ago

Which projectiles will be visibly affected by air resistance when they fall?

NemiM [27]

Any object that is launched as a projectile will lose speed and, as a result, altitude, as it travels through the air. The rate at which the object loses speed and altitude depends on the amount of force that way applied to it when it was launched. It is also dependent on the size and shape of the item. This is why something like, say, a football is much faster to fall to the ground than a bullet.

4 0

3 years ago

A roundabout in a fairground requires an input power of 2.5 kW when operating at a constant angular velocity of 0.47 rad s–1 . (

natita [175]

Answer:

Explanation:

Since the roundabout is rotating with uniform velocity ,

input power = frictional power

frictional power = 2.5 kW

frictional torque x angular velocity = 2.5 kW

frictional torque x .47 = 2.5 kW

frictional torque = 2.5 / .47 kN .m

= 5.32 kN . m

= 5 kN.m

b )

When power is switched off , it will decelerate because of frictional torque .

5 0

3 years ago

the pressure of gas in a cylinder is 70 kilopascals. if the volume of the cylinder is reduced from 8.0 liters to 4.0 liters, wha

Olegator [25]

By Boyle's law:

P₁V₁ = P₂V₂

70*8 = P₂*4

P₂*4 = 70*8

P₂ = 70*8/4 = 140

P₂ = 140 kiloPascals.

3 0

3 years ago

Car A has a mass of 1,200 kg and is traveling at a rate of 22 km/hr. It collides with car B. Car B has a mass of 1,900 kg and is

anastassius [24]

The car A has a mass of 1200 kg.

The car B has the mass of 1900 kg.

It is given that velocity of car A is given as 22 Km/hr

The car B has the velocity of 25 Km/hr.

Let the mass of two bodies are denoted as $m_{1} \ and\ m_{2}$

Let the velocity of cars A and B are denoted as $v_{1} \ and\ v_{2}$

The momentum before collision is-

$p_{i} =m_{1} v_{1} +m_{2} v_{2}$

[Here p stand for momentum.]

We are asked to calculate the final momentum of the system after collision.

The answer of the question is based law of conservation of linear momentum.

As per law of conservation of linear momentum the sum total linear momentum for an isolated system is always constant.Hence irrespective of the type of collision[elastic and inelastic],the momentum of the system is always constant which is a universal truth.

Let after the collision the velocity of A and B are $v'_{1} \ and\ v'_{2}$

Hence the final momentum of the system is-

$p_{f} = m_{1} v'_{1} +m_{2} v'_{2}$

As per the law of conservation of linear momentum, the initial and final momentum must be equal i.e

$p_{i} =p_{f}$

$m_{1} v_{1} +m_{2}v_{2} =m_{1} v'_{1} +m_{2} v'_{2}$

Hence the option A is right.

7 0

3 years ago

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