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goldfiish [28.3K]
3 years ago
14

Body mechanics are the ways in which your bodies move during activity? True or false

Physics
1 answer:
leva [86]3 years ago
5 0

This is in fact true


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How much heat will be needed to warm 187 grams of water from 10 0C to 90 0C?
kvv77 [185]
<h3>Hello there!</h3>

Here, you are looking for the amount of heat put in for water, at a mass of 187 grams, to change by 80 degrees.

The equation commonly accepted to find the answer to questions like these is the specific heat formula.

The equation is Q = mc∆T, where Q is the amount of energy put in to raise the temperature by a certain amount, m is the mass, c is the specific heat capacity, and ΔT is the amount of temperature change.

The information given:

m = 187 grams

c = specific heat capacity of water, or in this case 1 calorie, or 4.184 joules (which is what we will be using)

ΔT = 80 degrees

Now just plug everything in to solve.

Q = 187 * 4.184 * 80

Q = 62592.64

So you have your answer: 62592.64 joules.

Hope this helped!

5 0
3 years ago
Number 10 and an explaination would be fabulous. thanks!
Sati [7]
Linear momentum has to be conserved. It was zero before the thread eas burned ... when nothing was moving ... so the momentum of the masses moving in opposite directions has to add up to zero. ... Momentum = mass times speed. ... In one direction, you have 5 kg times 1/5 m/s= 1 kg-m/s. ... We need 1 kg-m/s in the other direction. ... 7 kg times speed = 1 kg-m/s. ... Can you finish it from here ?
3 0
3 years ago
Read 2 more answers
An object of mass m moving at a speed of v1 possesses kinetic energy that is equal to KE1. When the object's speed is doubled, t
Vera_Pavlovna [14]

Answer:

KE₂ = 4KE₁

Explanation:

KE₁ = ½mv₁²

KE₂ = ½mv₂²

KE₂ = ½m(2v₁)²

KE₂ = 4(½mv₁²)

KE₂ = 4KE₁

7 0
3 years ago
A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.
iris [78.8K]

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

F = \frac{kq_1q_2}{d^2}

Here

k = Coulomb's Constant

q_{1,2} = Charge of each object

d = Distance

Our values are given as,

q_1 = 1 \mu C

q_2 = 6 \mu C

d = 1 m

k =  9*10^9 Nm^2/C^2

a) The electric force on charge q_2 is

F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}

F_{12} = 54 mN

Force is positive i.e. repulsive

b) As the force exerted on q_2 will be equal to that act on q_1,

F_{21} = F_{12}

F_{21} = 54 mN

Force is positive i.e. repulsive

c) If q_2 = -6 \mu C, a negative sign will be introduced into the expression above i.e.

F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}

F_{12} = F_{21} = -54 mN

Force is negative i.e. attractive

6 0
3 years ago
Car A starts in Sacramento at 11am. It travels along 400 mile route to Los Angeles at 60 mph. Car B starts from Los Angeles at n
damaskus [11]

Answer:

  • 38.89 miles

Explanation:

from the question we have the following:

distance between Sacramento and los angles = 400 miles

speed of car A = 60 mph

start time of car A = 11 am

speed of car B = 75 mph

start time of car B = 12 pm

distance of Fresno from Los Angeles = 150 miles

  • To start off let's allow car A to travel for one hour (from 11 am to 12 pm), during which it would have covered a distance of 60 miles.
  • Now the time would be 12 pm and the distance between the two cars would be 400 - 60 (distance traveled by car A within 11 am to 12 pm) = 340 miles
  • From 12 pm to the time both cars will meet, the distance covered by car A + distance covered by car B would be equal to 340 miles. Therefore
  • Distance covered by car A = speed x time(t) = 60 x t = 60t
  • Distance covered by car B = speed x time(t) = 75 x t = 75t
  • 60t + 75t = 340 miles
  • 135t = 340
  • t = 2.51 hours
  • Recall that at their meeting point, the distance covered by car B = 75t = 75 x 2.62 = 188.89 miles
  • Since Fresno is 150 miles from Los Angeles, car B which is 188.89 miles from Los Angeles at their meeting point would be 188.89 - 150 = 38.89 miles from Fresno
  • 38.89 miles would also be the distance of car A from Fresno since that is their meeting point.

4 0
3 years ago
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