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alexgriva [62]
2 years ago
8

A school bus has a mass of 18,200 kg. The bus moves at 13.5 m/s. How fast must a 0.142-kg baseball move in order to have the sam

e momentum as the bus?
Physics
1 answer:
nikitadnepr [17]2 years ago
3 0

Answer:

bus momentum

p_bus= m_bus x v_bus

=18,200 x 16.5

basball momentum

pball=mball x vball

=0.142 x v

p_bus = pball

18200 x 16.5 = 0.142 x v

v=(18200 x 16.5)/0.142

v is the answer for baseball

Explanation:

⚠️not my answer tryna be honest here⚠️

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trapecia [35]

Answer:

a ) \dot m = 351.49 kg/s

b)  \dot m_{actual} = 1046.15 kg/s

Explanation:

given data:

pressure ration rp = 12

inlet temperature = 300 K

TURBINE inlet temperature  = 1000 K

AT the end of isentropic process (compression) temperature is

\frac{T_2'}{T_1} = rp ^{\frac{\gamma -1}{\gamma}}

\frac{T_2'}{300} = 12^{\frac{1.4 -1}{1.4}}

T_2' = 610.181 K

AT the end of isentropic process (expansion) temperature is

\frac{T_3}{T_4'} = rp ^{\frac{\gamma -1}{\gamma}}

\frac{1000'}{T_4'} = 12^{\frac{1.4 -1}{1.4}}

T_4' = 491.66 K

isentropic work is given as

w(compressor) = CP (T_2' -T_1)

w = 1.005(610.18 - 300)

w = 311.73 kJ/kg

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w(turbine) = 510.88 kJ/kg

a) mass flow rate for isentropic process is given as

\dot m = \frac{70000}{510.88 - 311.73}

\dot m = 351.49 kg/s

b) actual mass flow rate uis given as

\dot m_{actual} = \frac{70000}{51.088\times 0.85 - \frac{311.73}{0.85}}

\dot m_{actual} = 1046.15 kg/s

6 0
3 years ago
A 0.5 μF and a 11 μF capacitors are connected in series. Then the pair are connected in parallel with a 1.5 μF capacitor. What i
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Answer:

C_{eq}=1.97\ \mu F

Explanation:

Given that,

Capacitance 1, C_1=0.5\ \mu F

Capacitance 2, C_2=11\ \mu F

Capacitance 3, C_3=1.5\ \mu F

C₁ and C₂ are connected in series. Their equivalent is given by :

\dfrac{1}{C'}=\dfrac{1}{C_1}+\dfrac{1}{C_2}

\dfrac{1}{C'}=\dfrac{1}{0.5}+\dfrac{1}{11}

C'=0.47\ \mu F

Now C' and C₃ are connected in parallel. So, the final equivalent capacitance is given by :

C_{eq}=C'+C_3

C_{eq}=0.47+1.5

C_{eq}=1.97\ \mu F

So, the equivalent capacitance of the combination is 1.97 micro farad. Hence, this is the required solution.

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3 years ago
When the car was stopped by the tree, its change in velocity during the collision was-6 meters/second. This change in
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Answer: -3 meters/second^2.

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A block of 200 g is attached to a light spring with a force constant of 5 N / m and freely in a horizontal plane vibrates. The m
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Answer:

m  200 g , T  0.250 s,E 2.00 J

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2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

 

2

2 2 2.00 0.178 mm  200 g , T  0.250 s,E 2.00 J

;

2 2 25.1 rad s

T 0.250

 

   

(a)

 

2 2

k m    0.200 kg 25.1 rad s 126 N m

(b)

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2

2 2 2.00 0.178 m

Explanation:

That is a reason

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