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3 years ago

A 35 n object is on a 25 degree incline. the force of friction up the incline is 8.0 N.

1 answer:

4 0

The downwards component of weight of the object would be 35sin25 degrees = 14.8

F=14.8 - 8 = 6.8 N                                         m= 35/9.8= 3.57 kg
F=mA
Therefore,
3.57A= 6.8 N
=> A= 6.8/3.57
=> A= 1.902 ms^-2

F(max)= (U)R                       where (U)= coefficient of friction
                                                   and R= Normal reaction force

R= 35cos25
   = 31.72 N
Since, F(max)= 8
          8= (U)* 31.72
      =>(U)= 8/31.72
      =>(U)= 0.25

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3 years ago

A ball is thrown vertically upwards. It returns 6s later. Calculate : (1) the greatest height reached by the ball, and (2) the i

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Answer:

greatest displacement = 44.1m

initial velocity= 29.4m/s

Explanation:

Greatest displacement

s=1/2at^2

= (9.8/2 ×9)m

= 44.1m

initial velocity

s=ut-1/2at^2

44.1= 3u -(1/2×9.8×9)

44.1=3u-44.1

3u=88.2

u=29.4m/s

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4 years ago

A sailboat travels a distance of 600 m in 40 seconds. What speed is it going?

Gekata [30.6K]

Answer:

15 miles /seconds

Explanation:

Distance = 600m

Time = 40 seconds

Speed=?

$speed = \frac{distance}{time} \\ speed = \frac{600}{40}$

Simplify

$\frac{600}{40} = \frac{60}{4} \\ = 15$

7 0

4 years ago

A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this

wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

$r=r_o\times A^{\frac{1}{3}}$

$r_o=1.25 \times 10^{-15} m$ = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

$r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}$

$r=4.6656\times 10^{-15} m=4.6656 fm$

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) $F=k\times \frac{q_1q_2}{a^2}$

k= $9\times 10^9 N m^2/C^2$ = Coulombs constant

$q_1,q_2$ = charges kept at distance 'a' from each other

F = electrostatic force between charges

$q_1=+1.602\times 10^{-19} C$

$q_2=+1.602\times 10^{-19} C$

Force of repulsion between two protons on opposite sides of the diameter

$a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m$

$F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}$

$F=2.6527 N$

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

6 0

3 years ago

A long, hollow, cylindrical conductor (inner radius 3.4 mm, outer radius 7.3 mm) carries a current of 36 A distributed uniformly

Elden [556K]

Answer:

a. $B= 9.45 \times10^{-3} T$

b. $B= 0.820 T$

c. $B= 0.0584 T$

Explanation:

First, look at the picture to understand the problem before to solve it.

a. d1 = 1.1 mm

Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

To solve the equations we have to convert all units to those of the international system. (mm→m)

$B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\$

μ0 is the constant of proportionality

μ0=4πX10^-7 N*s2/c^2

b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

$J= \frac {-I_{c}}{\pi(c^{2}-b^{2} ) }}$

$J=\frac{-36}{\pi(5.33\times10^{-5}-1.16\times10^{-5}) } =-274.80\times10^{3} A/m^{2}$

$B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\$

$B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T$

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

$B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T$

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

3 0

3 years ago