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11 days ago

The pail is rotated at a constant rate so it has the minimum speed at all points along its circular path. The water has mass m.

When the pail is at the bottom of the circle, what is the magnitude of the force exerted by the water on the bottom of the pail?

Physics

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A solid sphere is released from the top of a ramp that is at a height h1 = 2.30 m. It rolls down the ramp without slipping. The

Yuliya22 [3333]

Answer:

The horizontal distance d that the ball covers before it lands is 1.72 m.

Explanation:

Given,

Height of ramp $h_{1}=2.30\ m$

Height of bottom of ramp $h_{2}=1.69\ m$

Diameter = 0.17 m

We need to determine the horizontal distance d the ball travels before landing.

We need to calculate the time

Using the equation of motion

$h_{2}=ut+\dfrac{1}{2}gt^2$

$t=\sqrt{\dfrac{2h_{2}}{g}}$

$t=\sqrt{\dfrac{2\times1.69}{9.8}}$

$t=0.587\ sec$

Next, we can find the ball's velocity

Using the kinetic energy formula

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

$K.E=\dfrac{1}{2}mv^2+\dfrac{1}{2}\times(\dfrac{2}{5}mr^2)\times(\dfrac{v}{r})^2$

$K.E=\dfrac{7}{10}mv^2$

By applying the conservation of energy

$K.E=mg(h_{1}-h_{2})$

$\dfrac{7}{10}mv^2=mg(h_{1}-h_{2})$

$v^2=\dfrac{10}{7}\times g(h_{1}-h_{2})$

We substitute the values into the equation

$v=\sqrt{\dfrac{10\times9.8\times(2.30-1.69)}{7}}$

$v=2.922\ m/s$

Next, we determine the horizontal distance d the ball travels before landing

Using the distance formula

$d =vt$

Where. d = distance

t = time

v = velocity

We substitute the values into the formula

$d=2.922\times 0.587$

$d=1.72\ m$

Thus, the horizontal distance d that the ball travels before it lands is 1.72 m.

8 0

1 month ago

A truck collides with a car on horizontal ground. At one moment during the collision, the magnitude of the acceleration of the t

inna [3103]

Response:

The car's acceleration magnitude is 35.53 m/s²

Details:

Given;

acceleration of the truck, $a_t$ = 12.7 m/s²

mass of the truck, $m_t$ = 2490 kg

mass of the car, $m_c$ = 890 kg

let the acceleration of the car during the collision = $a_c$

Using Newton's third law of motion;

The force exerted by the truck equals the force exerted by the car.

The car's force acts in the opposite direction.

$F_c = -F_t\\\\m_ca_c = -m_t a_t\\\\a_c =- \frac{m_ta_t}{m_c} \\\\a_c = -\frac{2490 \times 12.7}{890} \\\\a_c = - 35.53 \ m/s^2$

Thus, the car's acceleration magnitude is 35.53 m/s²

3 0

2 months ago

In an isolated system, the total heat given off by warmer substances equals the total heat energy gained by cooler substances. N

Keith_Richards [3271]

Answer:

The temperature of the cooler object was nearly at room temperature. As a result, the system underwent minimal change

Explanation:

In a closed system with two objects at varying temperatures, heat energy typically flows from the hotter object to the cooler one. This transfer is more pronounced when there is a significant temperature disparity between the objects. Conversely, if the temperature difference is minor or negligible, the resulting change will be minimal.

3 0

2 months ago

A cylinder has 500 cm3 of water. After a mass of 100 grams of sand is poured into the cylinder and all air bubbles are removed b

Keith_Richards [3271]

Answer: SG = 2.67

The specific gravity for the sand is 2.67

Explanation:

Specific gravity is determined by the formula: density of the substance/density of water

Provided information;

Mass of sand m = 100g

The volume of sand equals the volume of water it displaces

Vs = 537.5cm^3 - 500 cm^3

Vs = 37.5cm^3

Calculating density of sand = m/Vs = 100g/37.5 cm^3

Ds = 2.67g/cm^3

Density of water Dw = 1.00 g/cm^3

Thus, the specific gravity of the sand can be expressed as

SG = Ds/Dw

SG = (2.67g/cm^3)/(1.00g/cm^3)

SG = 2.67

The specific gravity of the sand stands at 2.67

3 0

2 months ago