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3 years ago

A 37 kg object has an applied force of 85N [R] acting on it. The coefficient of

Physics

1 answer:

Sliva [168]3 years ago

6 0

Answer:

Explanation:

This is quite tricky! You need to do 2 different equations to solve all the parts of this problem. First is finding the acceleration in one dimension, which has an equation of

F - f = ma

where F is the applied Friction,

f is the frictional force acting against F,

m is the mass of the object, and

a is the acceleration of the object (NOT the velocity!)

This is Newton's Second Law expanded on a bit. The sum of the forces working on an object is equal to the object's mass times its acceleration. We have F, but we need f which is found in the equation

f = μ $F_n$ which is the coefficient of kinetic friction times the weight of the object. Weight is found in the equation

w = mg where m is mass and g is the pull of gravity. Let's start there and work backwards:

w = 37(9.8) to 2 sig figs so

w = 360N. Now fill that in to find f:

f = (.17)(360) to 2 sig figs so

f = 61. Now for the final answer in the original equation way back up at the top:

85 - 61 = 37a and do the subtraction on the left side first:

24 = 37a and then we divide to 2 sig figs to get

a = .65 m/s/s

Since we are moving in a straight line (as opposed to on an angle) the displacement is found in

d = rt which simply says that the distance an object moves is equal to its rate times the time. Therefore,

d = 2.2(3.4) to 2 sig figs so

d = 7.5 m

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A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$

$Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma \int ^R_0 * \delta (r-R) r^2 \ dr$

$Q = 4 \pi \sigma *R^2$ since $( \int ^{xo}_{0} (x -x_o) f(x) \ dx = f(x_o) )$

$\mathbf{Q = 4 \pi R^2 \sigma }$

6 0

3 years ago

If a ball is dropped from a height (H) its velocity will increase until it hits the ground (assuming that aerodynamic drag due

mario62 [17]

Answer:

9.801 m/s²

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 39 ft/s

s = Displacement = 720 cm = 7.2 m

a = Acceleration

Converting to m/s

$39\ ft/s=\frac{39}{3.281}=11.88\ m/s$

Equation of motion

$v^2-u^2=2as\\\Rightarrow a=\frac{v^2-u^2}{2s}\\\Rightarrow a=\frac{11.88^2-0^2}{2\times 7.2}\\\Rightarrow a=9.801\ m/s^2$

Acceleration of the ball is 9.801 m/s²

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3 years ago

If my science fair project is about if girls can see through illusions better than boys, what would my background research be ab

SCORPION-xisa [38]

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3 years ago

If a football player has more mass they will also have more ______ ?<br> Fill in the blank

serg [7]

If a football player has more mass, they will also have more <u>momentum</u>. This is because mass is directly proportional to momentum.

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3 years ago

Identify the magnetic North Pole of Earth's magnet

AlladinOne [14]

If i am reading this question right the answer should be the south pole

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3 years ago

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