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Genrish500 [490]
1 year ago
11

Two positive charge spheres, the spheres are separated by 0.40 m. The charge on the first sphere is 100 microcoulombs and the

Physics
1 answer:
agasfer [191]1 year ago
8 0

Answer: 168.75 N

Explanation:

first, let's convert microcoulombs to coulombs

q1 = 1e-4 C

q2 = 3e-5 C

r = 0.4 m

then use the equation Fe = \frac{kq_{1} q_{2}}{r^{2} }

plug in values --> F = (9e9*1e-4*3e-5)/(0.4)^2

F = 168.75 N

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What is the acceleration of a racing car if it's speed is increased uniformly from 44 m /s to 66 m / s over an 11 second period
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Answer:

the acceleration of the race car is 2 m/s²

Explanation:

Given;

initial velocity of the race car, u = 44 m/s

final velocity of the race car, v = 66 m/s

time of motion of the race car, t = 11 s

The acceleration of the race car is  calculated as;

a = \frac{v-u}{t} \\\\a = \frac{66-44}{11} \\\\a = 2 \ m/s^2

Therefore, the acceleration of the race car is 2 m/s²

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A sample of argon gas (molar mass 40 g) is at four times the absolute temperature of a sample of hydrogen gas (molar mass 2 g).
qaws [65]

To solve this problem, let us recall that the formula for gases assuming ideal behaviour is given as:

rms = sqrt (3 R T / M)

where

R = gas constant = 8.314 Pa m^3 / mol K

T = temperature

M = molar mass

Now we get the ratios of rms of Argon (1) to hydrogen (2):

rms1 / rms2 = sqrt (3 R T1 / M1) / sqrt (3 R T2 / M2)

or

rms1 / rms2 = sqrt ((T1 / M1) / (T2 / M2))

rms1 / rms2 = sqrt (T1 M2 / T2 M1)

Since T1 = 4 T2

rms1 / rms2 = sqrt (4 T2 M2 / T2 M1)

rms1 / rms2 = sqrt (4 M2 / M1)

and M2 = 2 while M1 = 40

rms1 / rms2 = sqrt (4 * 2 / 40)

rms1 / rms2 = 0.447

 

Therefore the ratio of rms is:

<span>rms_Argon / rms_Hydrogen = 0.45</span>

7 0
3 years ago
A cart loaded with bricks has a total mass of 9.13 kg and is pulled at constant speed by a rope. The rope is inclined at 24.7 ◦
blagie [28]

Answer:

W = 0.63 KJ

Explanation:

Work (W) is defined as the point product of force (F) by the distance (d)the body travels due to this force.  

W= F*d *cosα Formula (1)  

F : force (N)

d : displacement (m)

α : angle between force and displacement

Newton's second law:

∑F = m*a Formula (2)  

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration  (m/s²)

We define the x-axis in the direction parallel to the movement of the cart on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the cart

W: Weight of the cart : In vertical direction

FN : Normal force : perpendicular to the floor

f : Friction force: parallel to the floor

T : tension Force,  inclined at  θ=24.7° above the horizontal

Calculated of the W

W= m*g

W= 9.13 kg* 9.8 m/s² = 89.47 N

x-y components o the  tension force (T)

Tx = Tcosθ = T*cos 24.7° (N)

Ty = Tsin θ = T*sin 24.7°  (N)

Calculated of the FN  

We apply the formula (2)  

∑Fy = m*ay ay = 0  

FN +Ty- W = 0  

FN = W-Ty  

FN =  89.47-T*sin 24.7°

Calculated of the friction force (f)

f = μk*FN

f =(0.597)*(  89.47-T*sin 24.7° )

f= 53.41-0.249T

Calculated of the tension force of the rope (f)

We apply the formula (2) :

∑Fx = m*ax  ,  ax= 0 ,because the speed of the cart  is constant

Tx - f = 0

T*cos 24.7°-( 53.41 - 0.249T )= 0

T*cos 24.7° + 0.249T = 53.41

(1.1575)T = 53.41

T= (53.41) / (1.1575)

T= 46.14 N

Work done on the cart by the rope

We apply the formula (1)

W=T*d *cosα

W= (46.14 N)*(15.1 m) *(cos24.7)

W = 632.97 (N*m) = 632.97 (J)

W = 0.63 KJ

6 0
4 years ago
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