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1 year ago

Can you please help me with #6 (the 6 was wrong in the question it meant #5 as you can see I tried to change it)

Physics

1 answer:

Harman [31]1 year ago

5 0

$\begin{gathered} \text{Question 5)} \\ Q=52,000J \\ m=2kg \\ C=130\text{ J/Kg K} \\ \Delta T=\text{?} \\ Q=Cm\Delta T \\ \text{Solving }\Delta T \\ \Delta T=\frac{Q}{Cm} \\ \Delta T=\frac{52,000J}{(130\text{ J/Kg K)(}2kg\text{)}} \\ \Delta T=200K,\text{ but Kelving and }degree\text{ Celsius have the same change} \\ \text{Hence} \\ \Delta T=200\text{ \degree{}C} \\ \\ \text{Question 6} \\ T1=350\text{ K} \\ \Delta T=200\text{ K} \\ T2=\text{?} \\ \Delta T=T2-T1 \\ \text{Sol}v\text{ing T2} \\ T2=\Delta T+T1 \\ T2=200\text{ K}+350\text{ K} \\ T2=550\text{ K} \\ \text{The final temperature is }550\text{ K} \end{gathered}$

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When a guitar is tuned to adjust it pitch what is it that is changed?

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Answer:

B) Its frequency

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3 years ago

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stepan [7]

Answer: physical or mechanical weathering

Explanation:

Mechanical weathering which is also referred to as the physical weathering occurs when a rock is broken down into smaller pieces. In this case, there will be a physical change of the rock but its composition will not change.

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3 0

3 years ago

Mary and her younger brother Alex decide to ride the carousel at the State Fair. Mary sits on one of the horses in the outer sec

Alex

Answer:

ωM = 1.7 rad/s

vM = 3.4 m/s

ωA = 1.7 rad/s

vA = 1.7 m/s

Explanation:

Calculation of the angular speed (ω)

We use the following formula for circular motion:

ω = θ / t Formula (1)

Where:

ω: angular speed (rad/s)

θ:angle that the particle travels (rad)

t: time interval (s)

Equivalence

1 revolution = 2π rad

Data

θ= 1 revolution = 2π rad

t = 3.7 s

We replace data in the formula (1):

ω = (2π rad) / (3.7s )= 1.7 rad/s

Because the angular speed depends only on the number of turns in a time interval then Mary's angular speed (ωM) is equal to Alex's angular speed (ωA).

ωM = ωA = 1.7 rad/s

Calculation of the tangential speed (v)

We use the following formula for circular motion:

v = ω*r Formula (2)

Where:

v : tangential speed (m/s)

ω: angular speed ( rad/s)

r : radius of the circular path (m)

Data

ωM = ωM = ω= 1.7 rad/s

rM = 2m : Mary's circular path radius

rA = 1m : Alex's circular path radius

We apply the formula (2) to calculate v:

vM = ω* rM = (1.7) (2) = 3.4 m/s

vA = ω* rA = (1.7) (1) = 1.7 m/s

7 0

4 years ago

A car of mass 1000 kg is moving at 25 m/s. It collides with a car of mass 1200 kg moving at 30 m/s. When the cars collide, they

Alinara [238K]

Answer:

The total momentum of the cars before the collision is 61,000 kg.m/s

The total momentum of the cars after the collision is 61,000 kg.m/s

The velocity of the cars after the collision is 27.727 m/s

Explanation:

Given;

mass of the first car, m₁ = 1000 kg

initial velocity of the car, u₁ = 25 m/s

mass of the second car, m₂ = 1200 kg

initial velocity of the second car, u₂ = 30 m/s

The common velocity of the cars after collision = v

The total momentum of the cars before collision is calculated as;

P₁ = m₁u₁ + m₂u₂

P₁ = (1000 x 25) + (1200 x 30)

P₁ = 61,000 kg.m/s

The total momentum of the cars after collision is calculated as;

P₂ = m₁v + m₂v

where;

v is the common velocities of the cars after collision since they stick together.

P₂ = v(m₁ + m₂)

To determine "v" apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁ + m₂u₂ = v(m₁ + m₂)

(1000 x 25) + (1200 x 30) = v(1000 + 1200)

61,000 = 2,200v

v = 61,000/2,200

v = 27.727 m/s

The total momentum after collsion = v(m₁ + m₂)

= 27.727(1000 + 1200)

= 61,000 kg.m/s

Thus, momentum before and after collsion are equal.

8 0

3 years ago