This downward force pulls indoor skydivers towards the Earth. *

A 0.30-m-radius automobile tire accelerates from rest at a constant 2.0 rad/s2. What is the centripetal acceleration of a point

lianna [129]

Answer:

The centripetal acceleration is 30 m/s²

Explanation:

Given that,

Radius = 0.30 m

Acceleration = 2.0 rad/s²

Time = 5.0

We need to calculate the centripetal acceleration

Using formula of angular velocity

$\omega=a\times t$

Where,

a = acceleration

t = time

Put the value into the formula

$\omega=2.0\times5.0$

$\omega=10\ rad/s$

Using formula of centripetal acceleration

$a_{c}=r\omega^2$

$a_{c}=0.30\times10^2$

$a_{c}=30\ m/s^2$

Hence, The centripetal acceleration is 30 m/s²

8 0

3 years ago

Calculate the amount of heat needed to raise 1.0 kg of ice at -20 degrees Celsius to steam at 120 degree Celsius

CaHeK987 [17]

Answer:

801.1 kJ

Explanation:

The ice increases in temperature from -20 °C to 0 °C and then melts at 0 °C.

The heat required to raise the ice to 0 °C is Q₁ = mc₁Δθ₁ where m = mass of ice = 1 kg, c₁ = specific heat capacity of ice = 2108 J/kg°C and Δθ₁ = temperature change. Q₁ = 1 kg × 2108 J/kg°C × (0 - (-20))°C = 2108 J/kg°C × 20 °C = 4216 J

The latent heat required to melt the ice is Q₂ = mL₁ where L₁ = specific latent heat of fusion of ice = 336000 J/kg. Q₁ = 1 kg × 336000 J/kg = 336000 J

The heat required to raise the water to 100 °C is Q₃ = mc₂Δθ₂ where m = mass of ice = 1 kg, c₂ = specific heat capacity of water = 4187 J/kg°C and Δθ₂ = temperature change. Q₃ = 1 kg × 4187 J/kg°C × (100 - 0)°C = 4187 J/kg°C × 100 °C = 418700 J

The latent heat required to convert the water to steam is Q₄ = mL₂ where L = specific latent heat of vapourisation of water = 2260 J/kg. Q₄ = 1 kg × 2260 J/kg = 2260 J

The heat required to raise the steam to 120 °C is Q₅ = mc₃Δθ₃ where m = mass of ice = 1 kg, c₃ = specific heat capacity of steam = 1996 J/kg°C and Δθ₃ = temperature change. Q₃ = 1 kg × 1996 J/kg°C × (120 - 100)°C = 1996 J/kg°C × 20 °C = 39920 J

The total amount of heat Q = Q₁ + Q₂ + Q₃ + Q₄ + Q₅ = 4216 J + 336000 J

+ 418700 J + 2260 J + 39920 J = 801096 J ≅ 801.1 kJ

4 0

3 years ago

A projectile of mass 2.0 kg is fired in the air at an angle of 40.0 ° to the horizon at a speed of 50.0 m/s. At the highest poin

tekilochka [14]

Answer:

a) The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

0.7 kg is 109.4 ms on the x axis

b) Y = 109.3 m

Explanation:

This is a moment and projectile launch exercise.

a) Let's start by finding the initial velocity of the projectile

sin 40 = voy / v₀

$v_{oy}$ = v₀ sin 40

$v_{oy}$ = 50.0 sin40

$v_{oy}$ = 32.14 m / s

cos 40 = v₀ₓ / V₀

v₀ₓ = v₀ cos 40

v₀ₓ = 50.0 cos 40

v₀ₓ = 38.3 m / s

Let us define the system as the projectile formed t all fragments, for this system the moment is conserved in each axis

Let's write the amounts

Initial mass of the projectile M = 2.0 kg

Fragment mass 1 m₁ = 1.0 kg and its velocity is vₓ = 0 and $v_{y}$ = -10.0 m / s

Fragment mass 2 m₂ = 0.7 kg moves in the x direction

Fragment mass 3 m₃ = 0.3 kg moves up (y axis)

Moment before the break

X axis

p₀ₓ = m v₀ₓ

Y Axis y

$p_{oy}$ = 0

After the break

X axis

$p_{fx}$ = m₂ v₂

Axis y

$p_{fy}$ = m₁ v₁ + m₃ v₃

Let's write the conservation of the moment and calculate

Y Axis

0 = m₁ v₁ + m₃ v₃

Let's clear the speed of fragment 3

v₃ = - m₁ v₁ / m₃

v₃ = - (-10) 1 / 0.3

v₃ = 33.3 m / s

X axis

M v₀ₓ = m₂ v₂

v₂ = v₀ₓ M / m₂

v₂ = 38.3 2 / 0.7

v₂ = 109.4 m / s

The fragment speeds of 0.3 kg is 33.3 m / s on the y axis

0.7 kg is 109.4 ms on the x axis

b) The speed of the fragment is 33.3 m / s and has a starting height of where the fragmentation occurred, let's calculate with kinematics

$v_{fy}$ ² = $v_{oy}$ ² - 2 gy

0 = $v_{oy}$ ²-2gy

y = $v_{oy}$ ² / 2g

y = 32.14² / 2 9.8

y = 52.7 m

This is the height where the break occurs, which is the initial height for body movement of 0.3 kg

$v_{f}$ ² = $v_{y}$ ² - 2 g y₂

0 = $v_{y}$ ² - 2 g y₂

y₂ = $v_{y}$ ² / 2g

y₂ = 33.3²/2 9.8

y₂ = 56.58 m

Total body height is

Y = y + y₂

Y = 52.7 + 56.58

Y = 109.3 m

8 0

3 years ago

A wire is carrying current vertically downward. What is the direction of the force due to Earth's magnetic field on the wire?

andrey2020 [161]

Answer: The correct option is A ( horizontally towards the east)

Explanation:

Magnetic field is a region around a magnet or a current- carrying conductor, where a magnetic force is experienced. The magnetic effect of electric current was first discovered in the early 1820 by Oersted. Using a wire that had current flowing through it and a pivoted magnetic needle, he discovered that the direction of deflection depended on the direction of the current and whether the wire was above or below the needle.

From the way the needle turns when current when current carrying wire is held parallel to it, he therefore concluded that:

--> a current has magnetic field all round it,

--> the magnetic field is in a direction perpendicular to the current.

The above discovery was now modified in Fleming's left hand rule which states that when conductor carrying current is placed in a magnetic field, the conductor will experience a force perpendicular to both the field and the flow of current.

Therefore from the question, a vertical wire carrying current in DOWNWARD direction is placed in a HORIZONTAL magnetic field directed to the NORTH. The direction of the force on the wire is to the EAST.

4 0

3 years ago

Help me please, ;) I could use it

erastova [34]

Answer:

The solution(s) are in order with respect to the attachments

$2.613\:\cdot10^5$ Joules ; 5. Adding the same amount of heat to two different objects will produce the same increase in temperature ; 2. Same speed in both ; 2. A

Explanation:

Diagram 1 ( Liquid Nitrogen ) : So as you can see, we want our units in Joules here, and can therefore multiply the mass of gaseous nitrogen and the latent heat of liquid nitrogen, to cancel the units kg, and receive our solution - in terms of Joules. Let's do it.

q ( energy removed ) = mass of nitrogen $*$ latent heat of liquid nitrogen,

q = 1.3 kg $*$ 2.01 $*$ 10⁵ J / kg = $1.3\:\cdot \:2.01\:\cdot \:\:10^5$ = $10^5\cdot \:2.613$ = $100000\cdot \:2.613$ = $261300$ Joules = $261.3$ kiloJoules = 2.613 $*$ 10⁵Joules is the energy that must be removed

Diagram 2 : The same amount of heat does not necessarily mean the same increase in temperature for two different objects. The increase in temperature depends on the specific heat capacity of the substance. Therefore your solution is 5 ) Adding the same amount of heat to two different objects will produce the same increase in temperature.

Diagram 3 : The temperatures in both glasses are the same, and hence the molecules have the same average speed. Therefore your solution is 2 ) Same speed in both.

Diagram 4 : Glass A has more water molecules, and hence has more thermal energy. Your solution is 2 ) A.

7 0

3 years ago

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