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Stells [14]
3 years ago
12

Answer all of these questions and you will get the brainlist

Physics
2 answers:
adelina 88 [10]3 years ago
6 0

Answer:

1. kinetic energy = ½mv²

m= 2.1 kg

v = 30m/s

Ek = ½ ×2.1×30×30

= 945J

2. potential energy = mgh

m = 190kg

g = 9.8

h = 45m

Ep = 190×9.8 × 45

= 83790J

3.potential energy = mgh

m= 75kg

g= 9.8

h = 300m

Ep = 75×9.8×300

= 220,500J

Gnoma [55]3 years ago
5 0

Answer:

2 ans : the bell has 8550 J energy

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Four charges 7 × 10−9 C at (0 m, 0 m), −9 × 10−9 C at (3 m, 3 m), 7 × 10−9 C at (1 m, 3 m), and −8 × 10−9 C at (−3 m, 2 m), are
Ivanshal [37]

Answer:

Magnitude of the resulting force on the 7 nC charge at the origin:

Fn₁= 23.95*10⁻⁹ N

Explanation:

Look at the attached graphic:

Charges of positive signs exert repulsive forces on q₁ + and charges of negative signs exert attractive forces on q₁ +.

q₁ experiences three forces (F₂₁,F₃₁,F₄₁) and we calculate them with Coulomb's law:

F = (k*q₁*q)/(d)²

d_{12} = \sqrt{3^{2}+3^{2}  }  = \sqrt{18} m : distance from q₁ to q₂

(d₁₂)² = 18 m²

d_{13} =\sqrt{1^{2}+3^{2}  } = \sqrt{10} m  : distance from q₁ to q₃

(d₁₃)² = 10 m²

d_{14} =\sqrt{3^{2}+2^{2}  } = \sqrt{13} m  : distance from q₁ to q₄

(d₁₄)² = 13 m²

K=  8.98755 × 10⁹ N *m²/C²

q₁=  7*10⁻⁹C

k*q₁=8.98755*10⁹ *7*10⁻⁹= 62.9

F₂₁= (62.9)*(9* 10⁻⁹) /(18) = 31.45*10⁻⁹ C

F₃₁= (62.9)*(7* 10⁻⁹) /(10) = 44*10⁻⁹ C

F₄₁= (62.9)*(8* 10⁻⁹) /(13) = 38.7*10⁻⁹ C

x-y components of the net force on q₁ (Fn₁):

α= tan⁻¹(3/3)= 45°  ,  β= tan⁻¹(3/1)= 71.56° , θ= tan⁻¹(2/3)= 33.69°

Fn₁x = F₂₁x+ F₃₁x+F₄₁x

F₂₁x =+ F₂₁*cosα =+ (31.45*10⁻⁹)* (cos 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*cosβ = - ( 44*10⁻⁹)* (cos 71.56°) = -13.91 *10⁻⁹ N

F₄₁x= -F₄₁*cosθ = -(38.7*10⁻⁹)* (cos 33.69°) = -32.2*10⁻⁹ N

Fn₁x = (+22.24 - 13.91 - 32.2)*10⁻⁹ N

Fn₁x = -23.87 *10⁻⁹ N

Fn₁y = F₂₁y+ F₃₁y+F₄₁y

F₂₁x =+ F₂₁*sinα =+ (31.45*10⁻⁹)* (sin 45°) = +22.24 *10⁻⁹ N

F₃₁x= -F₃₁*sinβ = - ( 44*10⁻⁹)* (sin 71.56°) = -41.74 *10⁻⁹ N

F₄₁x= +F₄₁*sinθ = +(38.7*10⁻⁹)* (sin 33.69°) =+21.47*10⁻⁹ N

Fn₁y = (22.24 -41.74+21.47)*10⁻⁹ N  

Fn₁y = 1.97*10⁻⁹ N

Magnitude of the resulting force on the 7 nC charge at the origin (q₁):

F_{n1} =\sqrt{(Fn_{1x} )^{2}+(Fn_{1y} )^{2} }

F_{n1} =\sqrt{(23.87 )^{2}+(1.97 )^{2} }

Fn₁= 23.95*10⁻⁹ N

8 0
3 years ago
the police department determined that the force required to drag a 130 N(29 lb) car tire across the pavement at a constant veloc
ki77a [65]
The force of friction is given by:
f = μR, where μ is the friction coefficient and R is the reaction force, which will be equal to the weight.
100 = μ x 130
μ = 0.77
4 0
3 years ago
What is the formula for calculating the efficiency of a heat engine?
Anton [14]

Answer:

Efficiency = StartFraction T Subscript h Baseline minus T Subscript C Baseline over T Subscript h Baseline EndFraction times 100. Efficiency equals T Subscript c Baseline minus T Subscript h Baseline over T Subscript h Baseline all times 100.

7 0
3 years ago
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You wad up a piece of paper and throw it into the wastebasket. How far will
vitfil [10]

The range of the piece of paper is C) 1.4 m

Explanation:

The motion of the piece of paper is the motion of a projectile, which consists of two separate motions:

- A uniform motion along the horizontal direction, with constant velocity

- A uniformly accelerated motion along the vertical direction, with constant acceleration (the acceleration of gravity, g=9.8 m/s^2)

From the equation of motion, it is possible to find an expression for the range (the total horizontal distance covered) of a projectile, which is given by:

d=\frac{u^2 sin 2\theta}{g}

where

u is the initial velocity

\theta is the angle of projection

g is the acceleration of gravity

For the piece of paper in this problem,

u = 4.3 m/s

\theta=65^{\circ}

Substituting,

d=\frac{(4.3)^2 sin(2\cdot 65^{\circ})}{9.8}=1.45 m \sim 1.4 m

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

6 0
3 years ago
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AHH please help-
umka2103 [35]

Answer:

Radiation

Explanation:

I hope that help's

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3 years ago
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