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3 years ago

The more muscle mass a person builds, the more calories a person will burn. true or false

Physics

2 answers:

Oxana [17]3 years ago

8 0

False because as you are building up muscle
your weight stays the same

Lana71 [14]3 years ago

8 0

It definitely would be false

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A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release

Svetradugi [14.3K]

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m

a.Work done to compress the spring initially= $\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J$

By conservation law of energy

Initial energy of spring=Kinetic energy of object

$37.8=\frac{1}{2}(3)v^2$

$v^2=\frac{37.8\times 2}{3}$

$v=\sqrt{\frac{37.8\times 2}{3}}$

v=5.02 m/s

c.Work done by friction on the incline, $w_{friction}=P.E-spring \;energy$

$W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J$

8 0

3 years ago

Example of items that changed chemically

Ostrovityanka [42]

Burning of gases is one the example of chemical change

3 0

3 years ago

A bedroom bureau with a mass of 45 kg, including drawers and clothing, rests on the floor. (a) If the coefficient of static fric

sergij07 [2.7K]

Answer: the minimal force that you need to apply to move the bureau is F = 198.45N

Explanation:

If you want to move an object, you need to apply a force that is bigger than the force of the statical friction.

The force of statical friction can be written as.

Ff = k*N

where k is the coefficient of static friction, in this case, k = 0.45, and N is the normal force between the object and the surface.

In this case, the normal force is the weight of the bedroom bureau, this is:

N = m*g = 45kg*9.8m/s^2 = 441N

Then the force is:

Fr = 0.45*441N = 198.45N

This means that the minimal force that you need to apply to move the bureau is F = 198.45N

and after this point, the force of friction will work wit the kinetic coefficient of friction, that usually is smaller than the statical one.

4 0

3 years ago

You're driving down the highway late one night at 20 m/s when a deer steps onto the road 44 m in front of you. Your reaction tim

never [62]

Answer:

14.0 m

25.1 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Distance traveled in the reaction time

Distance = Speed × Time

$\text{Distance}=20\times 0.5=10\ m$

$v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-20^2}{2\times -10}\\\Rightarrow s=20\ m$

Distance in which the car will stop is 10+20 = 30.0 m

So, the car will not hit the deer

Distance between the car and deer is 44-30 = 14.0 m

$\text{Distance}=u\times 0.5=0.5u\ m$

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u^2=v^2-2as\\\Rightarrow u^2=0^2-2\times -10\times (44-0.5u)\\\Rightarrow u^2=20(44-0.5u)\\\Rightarrow u^2=880-10u\\\Rightarrow u^2+10u-880=0$

$u=\frac{-10+\sqrt{10^2-4\cdot \:1\left(-880\right)}}{2\cdot \:1}, u=\frac{-10-\sqrt{10^2-4\cdot \:1\left(-880\right)}}{2\cdot \:1}\\\Rightarrow u=25.08, -35.08\ m/s$