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2 years ago

I am primarily made of calcium. I protect and act as camouflage. I can be found as a filler in most household spices. What am I?

Physics

1 answer:

schepotkina [342]2 years ago

8 0

Answer:

Explanation:

Bones?

Bones are primarily made of calcium. Also, they protect the body, which is pretty much obvious. They also act as camouflage.

The only but I have is, can they be found as a filler in most household?

Asides that, bones pretty much fits in to the answer.

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The angle θ is slowly increased. Write an expression for the angle at which the block begins to move in terms of μs.

Reika [66]

Answer:

$tan \theta = \mu_s$

Explanation:

An object is at rest along a slope if the net force acting on it is zero. The equation of the forces along the direction parallel to the slope is:

$mg sin \theta - \mu_s R =0$ (1)

where

$mg sin \theta$ is the component of the weight parallel to the slope, with m being the mass of the object, g the acceleration of gravity, $\theta$ the angle of the slope

$\mu_s R$ is the frictional force, with $\mu_s$ being the coefficient of friction and R the normal reaction of the incline

The equation of the forces along the direction perpendicular to the slope is

$R-mg cos \theta = 0$

where

R is the normal reaction

$mg cos \theta$ is the component of the weight perpendicular to the slope

Solving for R,

$R=mg cos \theta$

And substituting into (1)

$mg sin \theta - \mu_s mg cos \theta = 0$

Re-arranging the equation,

$sin \theta = \mu_s cos \theta\\\rightarrow tan \theta = \mu_s$

This the condition at which the equilibrium holds: when the tangent of the angle becomes larger than the value of $\mu_s$ , the force of friction is no longer able to balance the component of the weight parallel to the slope, and so the object starts sliding down.

4 0

2 years ago

Ryan and Becca are moving a folding table out of the sunlight. A cup of lemonade, with the message 0.44 kg is on the table. Becc

jarptica [38.1K]

Calculate the weight of the table through the equation,

W = mg

where W is the weight, m is the mass, and g is the acceleration due to gravity. Substituting the known values,
W = (0.44 kg)(9.8 m/s²)
W = 4.312 N

The components of this weight can be calculated through the equation,

Wx = W(sin θ)

and Wy = W(cos θ)

x - component:
Wx = W(sin θ)
Substituting,
Wx = (4.312 N)(sin 150°) = 2.156 N

Wy = (4.312 N)(cos 150°) = -3.734 N

6 0

2 years ago

Suppose that you have left a 200-mL cup of coffee sitting until it has cooled to 30∘C , which you find totally unacceptable. You

AleksAgata [21]

Answer:

$\eta=0.5074\ or\ 50.74\%$

Explanation:

Considering the density & specific heat capacity of coffee to be equal to that of water.

GIVEN:

density $\rho=1\ g.mL^{-1}$

specific heat $c=4.186\ J.g^{-1}.K^{-1}$

mass of coffee, $m=200\times 1=200\ g$

initial temperature of coffee, $T_i=30^{\circ}C$

final temperature of coffee, $T_f=60^{\circ}C$

power rating of oven, $P=1100\ W$

time taken to reach the final temperature, $t=35\ s$

Heat released by the coffee to come to 60°C:

$Q=m.c.\Delta T$

$Q=200\times 4.186\times 30$

$Q=[tex]\eta=\frac{25116}{49500}$ \ J[/tex]

Now the energy used by the oven in the given time:

$E=P.t$

$E=1100\times 45$

$E=49500\ J$

Now the efficiency:

$\eta=\frac{Q}{E}$

$\eta=0.5074\ or\ 50.74\%$

8 0

2 years ago

Describe the energy transformations that take place when a skier starts skiing down a hill but after a time is brought to rest b

Andrews [41]

The skier will transform their gravitational energy into mostly kinetic energy (with a minor amount transformed into heat from the friction of the skis across the snow and air friction). Once the skier hits the snowdrift, their kinetic energy is transferred into the snow which moves when they strike it due to the kinetic energy that is now in the snow. Along with again a minor amount of heat energy transferred as they move through the snowdrift.

6 0

2 years ago

a block of mass m slides along a frictionless track with speed vm. It collides with a stationary block of mass M. Find an expres

shusha [124]

Answer:

Part a) When collision is perfectly inelastic

$v_m = \frac{m + M}{m} \sqrt{5Rg}$