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alexira [117]
3 years ago
7

You want to use a rope to pull a 10-kg box of books up a plane inclined 30∘ above the horizontal. The coefficient of kinetic fri

ction is 0.29. The rope pulls parallel to the incline.
What force do you need to exert on the rope if you want to pull the box with a constant acceleration of 0.50 m/s2 up the plane?

Physics
1 answer:
fgiga [73]3 years ago
5 0

Answer:   78.61N

Explanation

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A carton is given a push across a horizontal, frictionless surface. The carton has a mass m, the push gives it an initial speed
sammy [17]

Answer and Explanation:

Data provided in the question

Carbon mass = m

Initial speed = v_i

Coefficient = μk

Based on the above information, the expressions are as follows

a. By using the energy considerations the expression for the carton moving distance is

As we know that

Fd = \frac{1}{2} m (v_i^2- v_f^2)

where,

v_f = 0

F = u_kmg

(\mu_kg) d = \frac{1}{2} m v_i^2

d = \frac{\frac{1}{2}v_i^2}{\mu_kg}

d = \frac{v_i^2}{2 \mu_kg}

b. The initial speed of the carton if the factor of 3 risen, so the expression is

v_i^1 = 3v_i

d^i = \frac{(3v_i^2)}{2\mu_kg}

= \frac{9v_i^2}{2\mu_kg}

d^i = 9(d)

8 0
3 years ago
A rough estimate of the radius of a nucleus is provided by the formula r 5 kA1/3, where k is approximately 1.3 × 10213 cm and A
Sphinxa [80]

Answer:

Density of 127 I = \rm 1.79\times 10^{14}\ g/cm^3.

Also, \rm Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.

Explanation:

Given, the radius of a nucleus is given as

\rm r=kA^{1/3}.

where,

  • \rm k = 1.3\times 10^{-13} cm.
  • A is the mass number of the nucleus.

The density of the nucleus is defined as the mass of the nucleus M per unit volume V.

\rm \rho = \dfrac{M}{V}=\dfrac{M}{\dfrac 43 \pi r^3}=\dfrac{M}{\dfrac 43 \pi (kA^{1/3})^3}=\dfrac{M}{\dfrac 43 \pi k^3A}.

For the nucleus 127 I,

Mass, M = \rm 2.1\times 10^{-22}\ g.

Mass number, A = 127.

Therefore, the density of the 127 I nucleus is given by

\rm \rho = \dfrac{2.1\times 10^{-22}\ g}{\dfrac 43 \times \pi \times (1.3\times 10^{-13})^3\times 127}=1.79\times 10^{14}\ g/cm^3.

On comparing with the density of the solid iodine,

\rm \dfrac{Density\ of\ ^{127}I}{Density\ of\ the\ solid\ iodine}=\dfrac{1.79\times 10^{14}\ g/cm^3}{4.93\ g/cm^3}=3.63\times 10^{13}.\\\\\Rightarrow Density\ of\ ^{127}I=3.63\times 10^{13}\times Density\ of\ the\ solid\ iodine.

7 0
3 years ago
If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle
ad-work [718]

This question is incomplete, the complete question is;

The electric force due to a uniform external electric field causes a torque of magnitude 20.0 × 10⁻⁹ N⋅m on an electric dipole oriented at 30° from the direction of the external field. The dipole moment of the dipole is 7.5 × 10⁻¹² C⋅m.

What is the magnitude of the external electric field?

If the two particles that make up the dipole are 2.5 mm apart, what is the magnitude of the charge on each particle?

Answer:

- the magnitude of the external electric field is 5333.3 N/C

- the magnitude of the charge on each particle is 3.0 × 10⁻¹² C  ≈ 3 nC

Explanation:

Given that;

Torque = 20.0 × 10⁻⁹ N⋅m

dipole moment = 7.5 × 10⁻¹²

∅ = 30°

The moment T of restoring couple is;

T = PEsin∅

E = T/Psin∅

we substitute

E = 20.0 × 10⁻⁹ N⋅m / (7.5 × 10⁻¹²) sin(30°)

E = 20.0 × 10⁻⁹ / 3.75 × 10⁻¹²

E =  5333.3 N/C

Therefore, the magnitude of the external electric field is 5333.3 N/C

The dipole moment is given by the expression;

p = ql

q = p / l

given that l = 2.5 mm = 0.0025 m

we substitute

q = 7.5 × 10⁻¹² / 0.0025

q = 3.0 × 10⁻¹² C ≈ 3 nC

Therefore, the magnitude of the charge on each particle is 3.0 × 10⁻¹² C ≈ 3 nC

7 0
3 years ago
A 1.0 m string with a 5 g stopper on the end is whirled in a vertical circle. The speed of the stopper is 8 m/s at the top of th
Andrew [12]

Answer:

Explanation:

A )

At the bottom of the circle , the potential energy of the stopper is converted into kinetic energy

1/2 m V² = mg x 2r + 1/2 mv²

m is mass of stopper , V is velocity at the bottom , r is radius of the circular path which is length of the string , v is velocity at the top

1/2  V² = g x 2r + 1/2 v²

 V² = g x 4r +  v²

 V² = 9.8 x 4 +  8²

V² = 103.2

V = 10.16 m/s

B )

If T be the tension at the top

Net downward force

= mg + T . This force provides centripetal force for the circular motion

mg +T = mv² / r

T =   mv²/r -mg

= m ( v²/r - g )

= .005 ( 8²/1 -g )

= .005 x 54.2

= .27 N .

C ) At the bottom

Net force = T  - mg , T is tension at the bottom , V is velocity at bottom

T-mg = mV²/r

T = m ( V²/r +g )

= .005 ( 10.16²/1 +9.8)

= .005 x 113

= .56 N .

3 0
3 years ago
Two cars collide at an intersection. Car A , with a mass of 2000 kg, is going from west to east, while car B, of mass 1500 kg, i
devlian [24]

Answer:

The answer of the part (a) is v2 = 7.09 m/s

and the answer of the part (b) is vA1 = 5.25 m/s

Explanation:

Explanation of the both parts of answer  is in the following attachments

6 0
3 years ago
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