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lawyer [7]
3 years ago
5

A lamp hangs vertically from a cord in a descending elevator that decelerates at 1.4 m/s2. (a) If the tension in the cord is 91

N, what is the lamp's mass
Physics
1 answer:
yanalaym [24]3 years ago
6 0

Answer:8.125 kg

Explanation:

Given

deceleration of elevator a=1.4\ m/s^2

suppose T is the tension in the cord

therefore net force on lamp is

F_{net}=T-mg

F_{net}=ma

direction of a and g is in the same direction

T-mg=m(1.4)

T=m(g+a)

m=\dfrac{T}{g+a}

m=\dfrac{91}{9.8+1.4}

m=8.125\ kg

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In a double slit experiment, 450 nm light passes through two slits producing an interference pattern where the first bright frin
musickatia [10]

Answer:

1.3 x 10⁻⁴ m

Explanation:

\lambda = wavelength of the light = 450 nm = 450 x 10⁻⁹ m

n = order of the bright fringe = 1

θ = angle = 0.2°

d = separation between the slits

For bright fringe, Using the equation

d Sinθ = n \lambda

Inserting the values

d Sin0.2° = (1) (450 x 10⁻⁹)

d (0.003491) = (450 x 10⁻⁹)

d = 1.3 x 10⁻⁴ m

6 0
2 years ago
If riding a lawnmower engine exerts 19 hp in one minute to move the lawnmower how much work is done
ioda

Answer:

the work done by the lawnmower is 236.14 J.

Explanation:

Given;

power exerted by the lawnmower engine, P = 19 hp

time in which the power was exerted, t = 1 minute = 60 s.

1 hp = 745.7 watts

The work done by the lawnmower is calculated as follows;

Work = Energy = \frac{Power}{time} \\\\Work = \frac{(19 \times 745.7)}{60} = 236.14 \ J

Therefore, the work done by the lawnmower is 236.14 J.

6 0
3 years ago
Which one of the following concepts explains why heavy nuclei do not follow the N = Z line (or trend) in the figure? A Transmuta
NikAS [45]

Answer:

Option B. Coulomb Repulsion

Explanation:

The reason for the heavy nucleus not being able to follow the trend or the N= Z line is because of the fact that as the atomic number, Z of an atom increases, the number of protons inside the atom also increases. Since the neutrons are charge less particles whereas the protons are positively charged particle and hence as these increases in number there is an increase in the repulsive force between the like charges, i.e., positively charged protons which is Coulomb repulsion.

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Calcular la cantidad de NOx y HC en volumen y otros puntos en la imagen sale
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3 years ago
Read 2 more answers
A spherical bowling ball with mass m = 3.6 kg and radius R = 0.101 m is thrown down the lane with an initial speed of v = 8.7 m/
luda_lava [24]

Answer:

1)  The magnitude of the angular acceleration = 67.92 rad/s^{2}

2) Magnitude of the linear acceleration = 2.744 m/s^{2}

3) How long does it take the bowling ball to begin rolling without slipping = 0.906 s

4) How long does it take the bowling ball to begin rolling without slipping = 6.75 m

5) the final velocity is 6.21 m/s

Explanation:

the given information :

Bowling mass m = 3.6 kg

Radius = 0.101 m

Initial speed v_{0} = 8.7 m/s

Coefficient of kinetic friction μ = 0.28

1) he magnitude of the angular acceleration of the bowling ball is

F = m a

F_{g}  = μ N  ,   N = m g

F_{g}  = μ m g

1) The magnitude of the angular acceleration of the bowling ball as it slides down the lane:

momen inersia of Bowling ball I = (2/5) m R^{2}

torque τ = I α

τ = F R

I α = F R

(2/5) m R^{2}  α = μ m g R

α = (5 μ g / 2R) μ g R

  = (5 x 0.28 x 9.8/ 2 x 0.101)

  = 67.92 rad/s^{2}

2) Magnitude of the linear acceleration of the bowling ball as it slides down the lane

F = - F_{g} , F_{k} is the force of kinetic friction

m a = - μ m g, remove m

the magnitude of linear accelaration is

a = μ g

  = (0.28) (9.8)

  = 2.744 m/s^{2}

3) The bowling ball takes time to begin rolling without slipping:

The linear speed, v_{t} = v_{0} - a t

                            v_{t}  =  v_{0} - μ g t

the angular speed, ω = ω0 + α t

                                ω = ω0 + (5  μ g/2R ) t

v_{t} = ω R

v_{0} - μ g t = ω0 R + (5  μ g/2R ) t R

7 μ g t/2 = v_{0} + ω0 R

hence,

t = (2 v_{0} + ω0 R)/  7 μ g

ω0 = 0 (no initial spin), therefore

t = 2 v_{0} / 7 μ g

 = 2 x 8.7 / 7 (0.28) (9.8)

 = 0.906 s

4) How long it takes for the bowling ball to begin rolling without slipping, S

S = v_{0}  t - (1/2) a t^{2}

  = (8.7) (0.906) - (1/2) (2.744) 0.906^{2}

  = 6.75 m

5) The final velocity

v_{t} = v_{0} - a t

v_{t} = 8.7 - (2.744) (0.906)

v_{t} = 6.21 m/s

4 0
3 years ago
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