Ask question

Ask question

All categories

2 years ago

11

A 95-kg astronaut is stranded from his space shuttle. He throws a 2-kg hammer away from the shuttle with a velocity of 19 m/s .

How fast will he be propelled toward the shuttle, in m/s ? (Round your answer to one decimal place if necessary )

1 answer:

geniusboy [140]2 years ago

5 0

Answer:

The astronaut will be propelled towards the shuttle at the rate of 0.4 m/s.

Explanation:

Given;

mass of the astronaut, m₁ = 95 kg

mass of the hammer thrown, m₂ = 2 kg

velocity of the hammer, v₂ = 19 m/s

let the recoil velocity of the shuttle = v₁

Apply the principle of conservation of linear momentum;

m₁v₁ = m₂v₂

v₁ = m₂v₂/m₁

v₁ = (2 x 19) / 95

v₁ = 0.4 m/s

Therefore, the astronaut will be propelled towards the shuttle at the rate of 0.4 m/s.

You might be interested in

Is kirchoff's law applicable for ac circuits?

SIZIF [17.4K]

<span>Kirchhoff's laws apply to AC circuits to either two cases: instantaneousneous values of currents and voltages or to complex values of currents and voltages. However, this never applies to: rms values of currents and voltages. Kirchoff's law relates to the current, voltage and resistance to multiple nodes</span>

4 0

3 years ago

What is the momentum of a child and a bicycle if the total mass of th

Sveta_85 [38]

bobo mag isip ayaw mag aral bobi

Explanation:

bobo ka boboboboboob

6 0

2 years ago

PLEASE HELP IM DESPERATE- what is the same and different about these two parabolas

krok68 [10]

One is bigger than the other and they are the same shape

7 0

2 years ago

A cellist tunes the C string of her instrument to a fundamental frequency of 65.4 Hz. The vibrating portion of the string is 0.5

Dahasolnce [82]

Answer:

a

$\lambda = 1.18 \ m$

b

$v = 77.172 \ m/s$

c

$T = 151.41 \ N$

Explanation:

From the question we are told that

The frequency is $f = 65.4 \ Hz$

The length of the vibrating string is $L = 0.590 \ m$

The mass is $m = 15.0 \ g = 0.015 \ kg$

Generally the wavelength is mathematically represented as

$\lambda = 2 * L$

=> $\lambda = 2 * 0.590$

=> $\lambda = 1.18 \ m$

Generally the wave speed is

$v = \lambda * f$

=> $v = 1.18 * 65.4$

=> $v = 77.172 \ m/s$

Generally the tension on the wire is mathematically represented as

$T = v^2 * \frac{ m }{L }$

=> $T = 77.172 ^2 * \frac{ 0.015 }{0.590}$

=> $T = 151.41 \ N$

7 0

3 years ago

Two blocks with masses 1 and 2 are connected by a massless string that passes over a massless pulley as shown. 1 has a mass of 2

Bess [88]

Answer:

The acceleration of $M_2$ is $a = 0.7156 m/s^2$

Explanation:

From the question we are told that

The mass of first block is $M_1 = 2.25 \ kg$

The angle of inclination of first block is $\theta _1 = 43.5^o$

The coefficient of kinetic friction of the first block is $\mu_1 = 0.205$

The mass of the second block is $M_2 = 5.45 \ kg$

The angle of inclination of the second block is $\theta _2 = 32.5^o$

The coefficient of kinetic friction of the second block is $\mu _2 = 0.105$

The acceleration of $M_1 \ and\ M_2$ are same

The force acting on the mass $M_1$ is mathematically represented as

$F_1 = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

=> $M_1 a = T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1$

Where T is the tension on the rope

The force acting on the mass $M_2$ is mathematically represented as

$F_2 = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

$M_2 a = M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

At equilibrium

$F_1 = F_2$

So

$T - M_1gsin \theta_1 - \mu_1 M_1 g cos\theta_1 =M_2gsin \theta_2 - T -\mu_2 M_2 g cos\theta_2$

making a the subject of the formula

$a = \frac{M_2 g sin \theta_2 - M_1 g sin \theta_1 - \mu_1 M_1g cos \theta - \mu_2 M_2 g cos \theta_2 }{M_1 +M_2}$

substituting values $a = \frac{(5.45) (9.8) sin (32.5) - (2.25) (9.8) sin (43.5) - (0.205)*(2.25) *9.8cos (43.5) - (0.105)*(5.45) *(9.8) cos(32.5) }{2.25 +5.45}$

=> $a = 0.7156 m/s^2$

3 0

3 years ago