Answer:
Explanation:
Given that,
Mass of star M(star) = 1.99×10^30kg
Gravitational constant G
G = 6.67×10^−11 N⋅m²/kg²
Diameter d = 25km
d = 25,000m
R = d/2 = 25,000/2
R = 12,500m
Weight w = 690N
Then, the person mass which is constant can be determined using
W =mg
m = W/g
m = 690/9.81
m = 70.34kg
The acceleration due to gravity on the surface of the neutron star is can be determined using
g(star) = GM(star)/R²
g(star) = 6.67×10^-11 × 1.99×10^30 / 12500²
g (star) = 8.49 × 10¹¹ m/s²
Then, the person weight on neutron star is
W = mg
Mass is constant, m = 70.34kg
W = 70.34 × 8.49 × 10¹¹
W = 5.98 × 10¹³ N
The weight of the person on neutron star is 5.98 × 10¹³ N
Answer:
The current pass the 
is 
Explanation:
The diagram for this question is shown on the first uploaded image
From the question we are told that
The voltage is 
The first resistance is 
The second resistance is 
Since the resistors are connected in series their equivalent resistance is
Substituting values
Since the resistance are connected in serie the current passing through the circuit is the same current passing through which is mathematically evaluated as
Substituting values
Responder:
192,900.64 Julios
Explicación:
La energía cinética es la energía que posee un cuerpo en virtud de su movimiento. Sea la masa del cuerpo m, su velocidad v.
Energía cinética = 1/2 mv²
Parámetros dados
masa del coche = 500 kg
velocidad = 100 km / h
La velocidad debe estar en m / s según la unidad internacional estándar. Al convertir;
100 km / h = 100 * 1000/3600 m / s
100 km / h = 27,8 m / s
velocidad del cuerpo = 27,8 m / s
Necesario
Energía cinética del coche = 1/2 * 500 * 27,8²
Energía cinética del automóvil = 1/2 * 500 * 771.6
Energía cinética del automóvil = 1/2 * 385,801.28
Energía cinética del automóvil = 192,900.64 Julios
<em>Por lo tanto, la energía cinética del automóvil es 192,900.64 julios.</em>
Answer:
1) 3.07kgm/s
2) 5.56kgm/s
3) 76.16N
4) 4.33kgm/s
5) 0.57s
6) -8.66J
Explanation:
Given
m = 0.221kg
v = 13.9m/s
θ = 25°
t = 0.073s
1) to get the magnitude of the initial momentum is the racquet ball. We use the formula,
P(i) = mv(i)
P(i) = 0.221 * 13.9
P(i) = 3.07kgm/s
2) Magnitude of the change in momentum of the ball,
P(i,x) = P(i) cos θ
P(i,x) = 3.07 * cos25
P(i,x) = 3.07 * 0.9063
P(i,x) = 2.78
ΔP = 2P(i,x)
ΔP = 2 * 2.78 = 5.56kgm/s
3) magnitude of the average force exerted by the wall,
F(ave) = ΔP/Δt
F(ave) = 5.56/0.073
F(ave) = 76.16N
4) ΔP(z) = mv(f) - mv(i)
ΔP(z) = 0.221*-7.8 - 0.221*11.8
ΔP(z) = -1.72 - 2.61
ΔP(z) = 4.33kgm/s
5) F(ave) = ΔP/Δt
Δt = ΔP/F(ave)
Δt = 4.33 / 76.16
Δt = 0.57s
6) KE(i) = 0.5mv(i)²
KE(f) = 0.5mv(f)²
ΔKE = 0.5m[v(f)² - v(i)²]
ΔKE = 0.5 * 0.221 [(-7.8)² - 11.8²]
ΔKE = 0.1105 ( 60.84 - 139.24 )
ΔKE = 0.1105 * -78.4
ΔKE = -8.66J
