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DedPeter [7]
3 years ago
12

Boy pulls a 5.0-kg sled with a rope that makes a 60.0° angle with respect to the horizontal surface of a frozen pond. The boy pu

lls on the rope with a force of 10.0 N; and the sled moves with constant velocity. What is the coefficient of friction between the and the ice?
(a) 0.09
(b) 0.12
(c) 0.18
(d) 0.06
(e) 0.24
Physics
1 answer:
kykrilka [37]3 years ago
4 0

Answer:

0.1

Explanation:

mass, m = 5 kg

θ = 60°

Force, F = 10 N

velocity is constant , it means the net force is zero.

So, the component of force along the surface is equal to the friction force

FCosθ = friction force

10 x cos 60 = μ x m x g

where, μ is the coefficient of friction

5 = μ x 5 x 9.8

μ = 0.1

Thus, the coefficient of friction is 0.1

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eimsori [14]
Sprry o cant see the words clearly
7 0
3 years ago
n alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 5.94 cm in a uniform magnetic field with B = 1.10 T.
9966 [12]

Answer:

a). V = 3.13*10⁶ m/s

b). T = 1.19*10^-7s

c). K.E = 2.04*10⁵

d). V = 1.02*10⁵V

Explanation:

q = +2e

M = 4.0u

r = 5.94cm = 0.0594m

B = 1.10T

1u = 1.67 * 10^-27kg

M = 4.0 * 1.67*10^-27 = 6.68*10^-27kg

a). Centripetal force = magnetic force

Mv / r = qB

V = qBr / m

V = [(2 * 1.60*10^-19) * 1.10 * 0.0594] / 6.68*10^-27

V = 2.09088 * 10^-20 / 6.68 * 10^-27

V = 3.13*10⁶ m/s

b). Period of revolution.

T = 2Πr / v

T = (2*π*0.0594) / 3.13*10⁶

T = 1.19*10⁻⁷s

c). kinetic energy = ½mv²

K.E = ½ * 6.68*10^-27 * (3.13*10⁶)²

K.E = 3.27*10^-14J

1ev = 1.60*10^-19J

xeV = 3.27*10^-14J

X = 2.04*10⁵eV

K.E = 2.04*10⁵eV

d). K.E = qV

V = K / q

V = 2.04*10⁵ / (2eV).....2e-

V = 1.02*10⁵V

7 0
3 years ago
A 5.22×104 kg railroad car moves on frictionless horizontal rails until it hits a horizontal spring stopper with a force constan
In-s [12.5K]

To solve this problem we will apply the principles of conservation of energy, for which we have to preserve the initial kinetic energy as elastic potential energy at the end of the movement. If said equality is maintained then we can affirm that,

\text{Initial Energy}=\text{Final Energy}

\frac{1}{2} mv^2=\frac{1}{2} kx^2

Here,

m = mass

k = Spring constant

x = Displacement

v = Velocity

Rearranging to find the velocity,

mv^2 = kx^2

v^2 = \frac{kx^2}{m}

v = \sqrt{\frac{kx^2}{m}}

Our values are,

m = 5.22*10^4kg

k = 4.58*10^5N/m

x = 32cm = 0.32m

Replacing our values we have,

v = \sqrt{\frac{(4.58*10^5)(5.22*10^4)}{0.32}}

v = 2.733*10^5m/s

Therefore the velocity is 2.733*10^5m/s

8 0
3 years ago
If you quadruple the temperature of a black body, by what factor will the total energy radiated per second per square meter incr
Anastaziya [24]
Radiant heat transfer is proportional to the 4-th power of absolute temperature.
Therefore if the temperature is quadrupled, the radiant heat energy will increase by a factor of
4⁴ = 256

Answer: 256
8 0
3 years ago
Plz help me with this!
vagabundo [1.1K]
A conductor is something which allows electricity to pass through it
an example is a coin
An insulator is something which cannot allow electricity to pass through it
an example is plastic (or plastic cotton reel, plastic container ect.)
Static electricity is when you rub two things together to create static

I'm not sure if i helped on the last question but i am definitely sure of the first ones
7 0
3 years ago
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