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Galina-37 [17]
3 years ago
6

The downward acceleration of a falling body on Earth is 9.81m/s2. On the moon the same quantity is 1.62m/s2. An astronaut in a s

pace suit has a mass of 145kg. (a) What is the astronaut’s weight on Earth? (b) On the moon? (c) What is the astronaut’s mass on the moon?
Physics
1 answer:
Bas_tet [7]3 years ago
5 0
The astronaut's mass doesn't change.  It's the same wherever he goes,
because it doesn't depend on what else is around him.

His weight depends on what else is near him, so it changes, depending
on where he is.

         Weight  =  (mass) x (gravity)

On Earth,  Weight = (145 kg) x (9.81 m/s²)  =  1,422.5 newtons.
                                                               (about 320 pounds)

On the moon, Weight = (145 kg) x (1.62 m/s²)  =  234.9 newtons.
                                                                   (about 53 pounds)
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PHYSICS<br> I need help with number two!!!!
EleoNora [17]

Answer:

mgh₁ + ½mv₁² = mgh₂ + ½mv₂²

Explanation:

Initial total energy = final total energy

PE₁ + KE₁ = PE₂ + KE₂

mgh₁ + ½mv₁² = mgh₂ + ½mv₂²

7 0
3 years ago
A coin 15.0 mm in diameter is placed 15.0 cm from a spherical mirror. The coin's image is 5.0 mm in diameter and is erect. Is th
s344n2d4d5 [400]

Answer:

15 cm

Explanation:

h_{o} = Diameter of the coin = 15 mm

h_{i} = Diameter of the image of coin =  5 mm

d_{o} = distance of the coin from mirror = 15 cm

d_{i} = distance of the image of coin from mirror = ?

Using the equation

\frac{d_{i}}{d_{o}} = \frac{- h_{i}}{h_{o}}

\frac{d_{i}}{15} = \frac{- (5)}{15}

d_{i} = - 5 cm

R = radius of curvature

Using the mirror equation

\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{2}{R}

\frac{1}{15} + \frac{1}{- 5} = \frac{2}{R}

R = - 15 cm

6 0
2 years ago
Suppose a wheel with a tire mounted on it is rotating at the constant rate of 3.33 times a second. A tack is stuck in the tire a
Troyanec [42]

Answer: 6.47m/s

Explanation:

The tangential speed can be defined in terms of linear speed. The linear speed is the distance traveled with respect to time taken. The tangential speed is basically, the linear speed across a circular path.

The time taken for 1 revolution is, 1/3.33 = 0.30s

velocity of the wheel = d/t

Since d is not given, we find d by using formula for the circumference of a circle. 2πr. Thus, V = 2πr/t

V = 2π * 0.309 / 0.3

V = 1.94/0.3

V = 6.47m/s

The tangential speed of the tack is 6.47m/s

7 0
3 years ago
Read 2 more answers
Una caja de 5.0kg de masa se acelera desde el reposo a través del piso mediante una fuerza a una tasa de 2.0 /s2 durante 7.0s en
Nady [450]

Responder:

<h2>490 julios </h2>

Explicación:

Se dice que el trabajo se realiza cuando una fuerza aplicada a un objeto hace que el objeto se mueva a través de una distancia. El trabajo realizado por un cuerpo se expresa mediante la fórmula;

Workdone = Fuerza * Distancia

Como Fuerza = masa * aceleración,

Workdone = masa * aceleración * distancia

Masa dada = 5.0kg, aceleración = 2.0m / s² d =?

Para obtener d, usaremos una de las leyes del movimiento,

d = ut + 1 / 2at²

u = 0 (ya que el cuerpo acelera desde el reposo) yt = 7.0s

d = 0 + 1/2 (2) (7) ²

d = 49m

Workdone = 5 * 2 * 49

Workdone = 490 Julios

4 0
3 years ago
A small airplane takes on 302 l of fuel. If the density of the fuel is 0.821 g/ml, what mass of fuel has the airplane taken on?
ella [17]

To calculate the mass of the fuel, we use the formula

m = V \times  \rho

Here, m is the mass of fuel, V is the volume of the fuel and its value is V =302 \ L =  302 \ L \times \frac{10^{3}m L }{L} = 302 \times 10^{3} \ mL and  \rho is the density and its value of  0.821 g/mL.

Substituting these values in above relation, we get  

m = 302 \times 10^{3} \ mL \times 0.821 g/mL = 247942 g \\\\ m = 247. 94 \ kg

Thus, the mass of the fuel 247 .94 kg.

8 0
2 years ago
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