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vitfil [10]
3 years ago
8

An astronaut in space cannot use a conventional means, such as a scale or balance, to determine the mass of an object. But she d

oes have devices to measure distance and time accurately. She knows her own mass is 78.4 kg, but she is unsure of the mass of a large gas canister in the airless rocket. When this canister is approaching her at 3.50 m/s, she pushes against it, which slows it down to 1.20 m/s (but does not reverse it) and gives her a speed of 2.40 m/s. What is the mass of this canister
Physics
1 answer:
Flura [38]3 years ago
7 0

Answer:

81.81 kg

Explanation:

m_1 = Mass of person = 78.4 kg

m_2 = Mass of canister

u_1 = Initial velocity of person = 0

u_2 = Initial velocity of canister = 3.5 m/s

v_1 = Final velocity of person = 2.4 m/s

v_2 = Final velocity of canister = 1.2 m/s

The momentum balance in the system is given by

m_1u_1+m_2u_2=m_1v_1+m_2v_2\\\Rightarrow 78.4\times 0+m_2\times 3.5=78.4\times 2.4+m_2\times 1.2\\\Rightarrow m_2=\dfrac{78.4\times 2.4}{3.5-1.2}\\\Rightarrow m_2=81.81\ \text{kg}

Mass of the canister is 81.81 kg.

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PLZ HELPPPPP!! i'll give brainliest
kati45 [8]

Answer:

12N to the right.

Explanation:

There is a force of 12N upwards and a force of 12N downwards: these cancel out.

Assign a negative value to forces towards the left, and a positive value to the forces towards the right: -3N and +15N

Combine them: -3N+15N = 12N

The net force has a magnitude of 12N, and since our answer was positive, it acts towards the right.

5 0
3 years ago
A 3.9 kg block is pushed along a horizontal floor by a force of magnitude 30 N at a downward angle θ = 40°. The coefficient of k
Luba_88 [7]

Answer:

The frictional force  F_{fri} = 6.446 N

The acceleration of the block a = 6.04 \frac{m}{s^{2} }

Explanation:

Mass of the block = 3.9 kg

\theta = 40°

\mu = 0.22

(a). The frictional force is given by

F_{fri} = \mu R_{N}

R_{N} = mg \cos \theta

R_{N} = 3.9 × 9.81 × \cos 40

R_{N} = 29.3 N

Therefore the frictional force

F_{fri} = 0.22 × 29.3

F_{fri} = 6.446 N

(b). Block acceleration is given by

F_{net} = F - F_{fri}

F = 30 N

F_{fri} = 6.446 N

F_{net} = 30 - 6.446

F_{net} = 23.554 N

The net force acting on the block is given by

F_{net}  = ma

23.554 = 3.9 × a

a = 6.04 \frac{m}{s^{2} }

This is the acceleration of the block.

8 0
3 years ago
The uniform rods AB and BC weigh 24 ky and kg, respectively,and the small wheel at C is of negligible weight. If the wheel ismov
victus00 [196]

The velocity of pin B after rod AB has rotated through 90* is vb = 3.2549 m/s.

<h3>What is Potential and Kinetic energy?</h3>

Potential energy is the energy that is stored in any item or system as a result of its location or component arrangement. The environment outside of the object or system, such as air or height, has no impact on it. In contrast, kinetic energy refers to the energy of moving particles inside a system or an item.

mass of rod, mab = 2.4kg

mass of rod, mbc = 4kg

conservation of energy

T_{1}  + V_{1} = T_{2}  + V_{2}

h_{ab}  = h_{bc}  = 0.18m

potential energy at position 1,

V1 = m_{ab} gh_{ab}  + m_{bc} gh_{bc}

V1 = 2.5 * 9.81 * 0.18 + 4 * 9.81 * 0.18

V1 = 11.30112

kinetic energy T1 at position 1 is zero

potential energy at position 2 is zero

K.E at position 2,

T_{2} = \frac{1}{2} l_{ab} w^{2}_{ab} +  \frac{1}{2} m_{bc} v^{2}_{G} +  \frac{1}{2} lw^{2}_{bc}

l_{ab} =\frac{m_{ab} l^{2}_{ab}  }{3}

= 1/3 *4 * (0.36)²

=0.10368kg m²

l =\frac{m_{bc} l^{2}_{bc}  }{12}

= 1/12 *4 * (0.6)²

=0.12kg m²

on putting the values in above equation we get,

T₂ = 1.0667vb²

0 + 11.30112 = 1.0667vb² + 0

vb = 3.2549 m/s

to learn more about Kinetic and potential energy go to - brainly.com/question/18963960

#SPJ4

5 0
1 year ago
Please Help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
arsen [322]

Answer:

Given that

speed u=4*10^6 m/s

electric field E=4*10^3 N/c

distance b/w the plates d=2 cm

basing on the concept of the electrostatices

now we find the acceleration b/w the plates  to find the horizontal distance traveled by the electron when it hits the plate.

acceleration a=qE/m=1.6*10^{-19}*4*10^3/9.1*10^{-31} =0.7*10^{15}=7*10^{14} m/s

now we find the horizontal distance traveled by electrons hit the plates

horizontal distance

X=u[2y/a]^{1/2}

=4*10^6[2*2*10^{-2}/7*10^{14}]^{1/2}

=3*10^{-2}= 3 cm

5 0
3 years ago
The graph shows the amplitude of a passing wave over time in seconds (s). What is the approximate frequency of the wave shown?
belka [17]

Answer:

B) 0.3Hz

Explanation:

I just took the test i hope i helped and i hope you pass the test

7 0
3 years ago
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