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2 years ago

7

A block of mass m sits at rest on a rough inclined ramp that makes an angle θ with the horizontal. What must be true about force

of static friction f on the block?

2 answers:

galina1969 [7]2 years ago

6 0

Answer:

μ =tanθ

Explanation:=

The ratio of the force of static friction and the normal reaction is equal to tanθ. F=mgsinθ. R = mgcosθ.

μ=tanθ

Aliun [14]2 years ago

5 0

Answer: So the static friction force must be equal to g*m*cos( θ)

Explanation: The force downwards of the block will be equal to the component with respect of where it can move.

So the force downwards will be:

F = Weight*cos( θ)

And we know that the weight is w = -m*g where g is the gravitational acceleration.

F = -g*m*cos( θ)

And we know that the block is at rest, so the net forces on it are equal to zero, this means that the gravitational force must be canceledby other one, and this can be the static friction force.

So the static friction force must be equal to g*m*cos( θ)

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An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electric

ELEN [110]

Answer:

a. 0 W b. ε²/R c. at R = r maximum power = ε²/4r d. For R = 2.00 Ω, P = 227.56 W. For R = 4.00 Ω, P = 256 W. For R = 6.00 Ω, P = 245.76 W

Explanation:

Here is the complete question

An external resistor with resistance R is connected to a battery that has emf ε and internal resistance r. Let P be the electrical power output of the source. By conservation of energy, P is equal to the power consumed by R. What is the value of P in the limit that R is (a) very small; (b) very large? (c) Show that the power output of the battery is a maximum when R = r . What is this maximum P in terms of ε and r? (d) A battery has ε= 64.0 V and r=4.00Ω. What is the power output of this battery when it is connected to a resistor R, for R=2.00Ω, R=4.00Ω, and R=6.00Ω? Are your results consistent with the general result that you derived in part (b)?

Solution

The power P consumed by external resistor R is P = I²R since current, I = ε/(R + r), and ε = e.m.f and r = internal resistance

P = ε²R/(R + r)²

a. when R is very small , R = 0 and P = ε²R/(R + r)² = ε² × 0/(0 + r)² = 0/r² = 0

b. When R is large, R >> r and R + r ⇒ R.

So, P = ε²R/(R + r)² = ε²R/R² = ε²/R

c. For maximum output, we differentiate P with respect to R

So dP/dR = d[ε²R/(R + r)²]/dr = -2ε²R/(R + r)³ + ε²/(R + r)². We then equate the expression to zero

dP/dR = 0

-2ε²R/(R + r)³ + ε²/(R + r)² = 0

-2ε²R/(R + r)³ = -ε²/(R + r)²

cancelling out the common variables

2R = R + r

2R - R = R = r

So for maximum power, R = r

So when R = r, P = ε²R/(R + r)² = ε²r/(r + r)² = ε²r/(2r)² = ε²/4r

d. ε = 64.0 V, r = 4.00 Ω

when R = 2.00 Ω, P = ε²R/(R + r)² = 64² × 2/(2 + 4)² = 227.56 W

when R = 4.00 Ω, P = ε²R/(R + r)² = 64² × 4/(4 + 4)² = 256 W

when R = 6.00 Ω, P = ε²R/(R + r)² = 64² × 6/(6 + 4)² = 245.76 W

The results are consistent with the results in part b

8 0

2 years ago

A hiker, caught in a rainstorm might absorb 1 liter of water in her clothing. if it is windy so that this amount of water is eva

NeTakaya

Water evaporates at 100⁰C

So change in temperature = 100-20 = 80⁰C

Amount of water to be evaporated = 1 liter = 1L*1kg/liter = 1 kg

Specific heat of water is 1 calorie/gram ⁰C = 4.186 joule/gram =4186 J/kg

So heat required E = mcΔT = 1 * 4186 *80= 334880 J =334.88 kJ

So amount of heat require to evaporate water = 334.88 kJ

4 0

2 years ago

the object is in _________ when the vector sum of the force is acting on the object is equal to zero

Norma-Jean [14]

Answer:

Equilibrium

Explanation:

An object is in equilibrium when the vector sum of the force acting on the object is equal to zero.

A body in equilibrium is at state of rest of rest or in motion with no external force acting on it.

The resultant of all forces acting on the body is zero.
In this case there is no net force and the body will be at rest.

7 0

2 years ago

PLEASE HELP ME WITH THIS PROBLEM

valentinak56 [21]

1) The mass of the continent is $2.13\cdot 10^{21} kg$

2) The kinetic energy of the continent is 274.8 J

3) The speed of the jogger must be 2.76 m/s

Explanation:

1)

The continent is a slab of side 5900 km (so the surface is 5900 x 5900, assuming it is a square) and depth 26 km, therefore its volume is:

$V=(36)(4600)^2=7.62\cdot 10^8 km^3 = 7.62\cdot 10^{17} m^3$

The mass of the continent is given by

$m=\rho V$

where:

$\rho = 2790 kg/m^3$ is its density

$V=7.62\cdot 10^{17} m^3$ is its volume

Substituting, we find the mass:

$m=(2790)(7.62\cdot 10^{17})=2.13\cdot 10^{21} kg$

2)

To find the kinetic energy, we need to convert the speed of the continent into m/s first.

The speed is

v = 1.6 cm/year

And we have:

1.6 cm = 0.016 m

$1 year = (365)(24)(60)(60)=3.15\cdot 10^7 s$

So, the speed is

$v=\frac{0.016 m}{3.15 \cdot 10^7 s}=5.08\cdot 10^{-10}m/s$

Now we can find the kinetic energy of the continent, which is given by

$K=\frac{1}{2}mv^2$

where

$m=2.13\cdot 10^{21} kg$ is the mass

$v=5.08\cdot 10^{-10}m/s$ is the speed

Substituting,

$K=\frac{1}{2}(2.13\cdot 10^{21})(5.08\cdot 10^{-10})^2=274.8 J$

3)

The jogger in this part has the same kinetic energy of the continent, so

K = 274.8 J

And its mass is

m = 72 kg

We can write his kinetic energy as

$K=\frac{1}{2}mv^2$

where

v is the speed of the man

And solving the equation for v, we find his speed:

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(274.8)}{72}}=2.76 m/s$

Learn more about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

3 0

2 years ago

A 150 g baseball is traveling horizontally at 50 m/s. If the ball takes 20 ms to stop once it is in contact with the catcher’s g

Sliva [168]

To solve for force, you need to get the product of mass and acceleration.
F = ma

Your given is:
m = 150g
a = ?
v = 50 m/s
t = 20ms

As you can see, you do not have acceleration yet. But if you read the problem you can come up with the formula of acceleration.
Acceleration is the change in velocity over a period of time.

a = change in velocity/time

To get the change in velocity, you get the difference between the initial velocity and final velocity:

$a = \frac{vf-vi}{t}$

The ball was moving initially at a velocity of 50 m/s and it came to a stop. This is your clue. If a ball comes to a stop then that means that the final velocity of the ball is 0 m/s.

So we can put it into our formula now:

$a = \frac{0m/s-50m/s}{20ms}$

WAIT! As you can see, the units do not match. We have ms and s into our equation and that means you cannot proceed till they are the same. First we need to convert ms to s.

$20ms$ x $\frac{1s}{1000ms}$ = $\frac{20s}{1000} = 0.02s$

So your new time is 0.02s. Now we put this time into the formula:

$a = \frac{0m/s-50m/s}{0.02s}$
$a = \frac{-50m/s}{0.02s} = -2,500 m/ s^{2}$

As you can see our acceleration is a negative value, this indicates that it decelerated or slowed down which makes sense because it was brought to a stop.

So now we have our acceleration. Now using this, we can get our force.

$F= ma$

Before we start doing this, you need to take note that the unit of force is N, but when you expand it, it is $kg.m/ s^{2}$ but as you can see our mass given is in grams. So again, before you put them into the equation we need to change it into kg first.

$150g = \frac{1kg}{1,000g} = \frac{150kg}{1,000} = 0.150kg$

Our new mass is 0.150kg.

To make things clearer, let us write down all our new values:

m = 0.150kg
a = -2,500 $m/ s^{2}$

Now that all our units match, we can put that into our formula:

$F= ma$
$F= (0.150kg)(-2,500m/s^{2})$
$F = -375kg.m/ s^{2} or -375N$

The value again is negative because it is going against the initial direction of the ball. But if your instructor just wants to get the value of force or the magnitude of the force, just disregard the sign.

4 0

2 years ago