What would the person’s mass be on the earth? Part B pls

When is the ball in a state of free fall?? Please help

anyanavicka [17]

The ball is in free fall when its coming towards the ground

7 0

3 years ago

Three liquids are at temperatures of 6 ◦C, 23◦C, and 38◦C, respectively. Equal masses of the first two liquids are mixed, and th

kupik [55]

The equilibrium temperature is T13=3.12 ◦C

Explanation:

Given

The temperature of liquids: T1=6◦C, T2=23◦C, T3=38◦C

The temperature of 1+2 liquids mix: T12= 13◦C.

The temperature of 2+3 liquids mix: T23=26.8 ◦C.

The temperature of 1+3 liquids mix: T13= ??

1.When the first two liquids are mixed:

mC1(T1-T12)+mC2(T2-T12)=0
C1(6-13)=C2(23-13)=0
7C1=10C2
C1=1.42C2

2.When the second and third liquids are mixed:

mC2(T2-T23)+mC3(T3-T23)=0
C2(23-26.8)=C3(38-26.8)=0
3.8C2=12.8C3
C2=3.36C3

3.When the first and third liquids are mixed:

mC1(T1-T13)+mC3(T3-T13)=0
C1(6-T13)+C3(38-T13)=0
C1=1.42C2 C2=3.36C3
C1=1.42C2(3.36C3)
C1=4.77C3
C1(6-T13)+C3(38-T13)=0
4.77C3(6-T13)+C3(38-T13)=0
By solving the equation we get,
T13=3.12 ◦C
The equilibrium temperature is T13=3.12 ◦C

7 0

3 years ago

If the weight of an object that is submerged in a fluid is 10 N and the buoyant force on it is

Nady [450]

Answer: the object should be overcome by buoyancy and rise in the fluid.

Explanation:

5 0

3 years ago

A large crate with mass m rests on a horizontal floor. The static and kinetic coefficients of friction between the crate and the

Mariana [72]

a) $F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}$

Here the crate is moving at constant velocity, so no acceleration:

a = 0

Let's analyze the forces acting along the horizontal and vertical direction.

- Vertical direction: the equation of the forces is

$R-Fsin \theta - mg = 0$ (1)

where

R is the normal reaction of the floor (upward)

$F sin \theta$ is the component of the force F in the vertical direction (downward)

mg is the weight of the crate (downward)

- Horizontal direction: the equation of the forces is

$F cos \theta - \mu_k R = 0$ (2)

where

$F cos \theta$ is the horizontal component of the force F (forward)

$\mu_k R$ is the force of friction (backward)

From (1) we get

$R=Fsin \theta +mg$

And substituting into (2)

$F cos \theta - \mu_k (Fsin \theta +mg) = 0\\F cos \theta -\mu _k F sin \theta = \mu_k mg\\F(cos \theta - \mu_k sin \theta) = \mu_k mg\\F=\frac{\mu_k mg}{cos \theta - \mu_k sin \theta}$

b) $\mu_s=cot(\theta)$

In this second case, the crate is still at rest, so we have to consider the static force of friction, not the kinetic one.

The equations of the forces will be:

$R-Fsin \theta - mg = 0$ (1)

$F cos \theta - \mu_s R = 0$ (2)

In this second case, we want to find the critical value of $\mu_s$ such that the woman cannot start the crate: this means that the force of friction must be at least equal to the component of the force pushing on the horizontal direction, $F cos \theta$ .

Therefore, using the same procedure as before,

$R=Fsin \theta +mg$

$F cos \theta - \mu_s (Fsin \theta +mg) = 0$

And solving for $\mu_s$ ,

$F cos \theta = \mu_s (Fsin \theta +mg) \\\mu_s = \frac{F cos \theta}{F sin \theta + mg}$

Now we analyze the expression that we found. We notice that if the force applied F is very large, $F sin \theta >> mg$ , therefore we can rewrite the expression as

$\mu_s \sim \frac{F cos \theta}{F sin \theta}\\\mu_s=cot(\theta)$

So, this is the critical value of the coefficient of static friction.

8 0

3 years ago

A 163 ‑turn circular coil of radius 2.65 cm and negligible resistance is immersed in a uniform magnetic field that is perpendicu

stiks02 [169]

Answer:

Explanation:

change in flux = no of turns x area of coil x change in magnetic field

= 163 x π 2.65² x 10⁻⁴ x ( .481 - 0 )

= 1728.84 x10⁻⁴ weber

rate of change of flux = change in flux / time

= 1728.84 x10⁻⁴ / .149

= 11603 x10⁻⁴ V .

volt induced = 1.16 V

current in the circuit

= volt induced / resistance

1.16 / 10.7 = .1084 A

energy dissipated = i²Rt where i is current , R is resistance and t is time.

= .1084² x 10.7 x .149

= .018734 J

= 18.73 mJ .

8 0

3 years ago