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Paraphin [41]
3 years ago
9

When a gun is fired how come yhe recoil is not strong enough to knock the shooter over?

Physics
1 answer:
ASHA 777 [7]3 years ago
8 0
Because inside the gun it has dynamite
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A car is stopped for a traffic signal. When the light turns green, the car accelerates, increasing its speed from 0 to 5.40 m/s
tekilochka [14]

Answer:

a) 378Ns

b) 477.27N

Explanation:

Impulse is the defined as the product of the applied force and time taken. This is expressed according to the formula

I = Ft = m(v-u)

m is the mass = 70kg

v is the final velocity = 5.4m/s

u is the initial velocity = 0m/s

Get the impulse

I = m(v-u)

I = 70(5.4-0)

I = 70(5.4)

I = 378Ns

b) Average total force is expressed as

F = ma (Newton's second law)

F = m(v-u)/t

F = 378/0.792

F = 477.27N

Hence the average total force experienced by a 70.0-kg passenger in the car during the time the car accelerates is 477.27N

3 0
3 years ago
Identifying Mechanical Energy
arsen [322]
Bruh ima be honest with you idek
3 0
3 years ago
Read 2 more answers
A vacuum cleaner produces sound with a measured sound level of 75.0 dB. (a) What is the intensity of this sound in W/m2? W/m2 (b
jonny [76]

Answer:

Part a)

I = 3.16 \times 10^{-5} W/m^2

Part b)

P_o = 0.162 Pa

Explanation:

Part a)

Level of sound = 75 dB

now we know that

L = 10 Log\frac{I}{I_0}

here we know that

I_0 = 10^{-12} W/m^2

now we have

75 = 10 Log(\frac{I}{10^{-12}})

I = 3.16 \times 10^{-5} W/m^2

Part b)

Intensity of sound wave is given as

I = \frac{1}{2}\rho A^2\omega^2 c

here we know that

A = \frac{P_o}{Bk}

so we have

I = \frac{1}{2}\rho(\frac{P_o}{Bk})^2\omega^2 c

I = \frac{1}{2}\rho P_o^2 \frac{c^3}{B^2}

now we know

\rho = 1.2 kg/m^3

c = 340 m/s

B = 1.4 \times 10^5 Pa

now we have

3.16 \times 10^{-5} = \frac{1}{2}(1.2)P_o^2\frac{340^3}{(1.4\times 10^5)^2}

P_o = 0.162 Pa

5 0
3 years ago
In which state of matter are water molecules measured as having a comparatively high temperature?
AnnZ [28]
Liquid water because if it said very high then it would be water vapor but it didn’t say that so the answer is B liquid water
8 0
3 years ago
A glass flask whose volume is 1000 cm^3 at a temperature of 1.00°C is completely filled with mercury at the same temperature. W
Marat540 [252]

Answer:

the coefficient of volume expansion of the glass is \mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

Explanation:

Given that:

Initial volume of the glass flask = 1000 cm³ = 10⁻³ m³

temperature of the glass flask and mercury= 1.00° C

After heat is applied ; the final temperature = 52.00° C

Temperature change ΔT = 52.00° C - 1.00° C = 51.00° C

Volume of the mercury overflow = 8.50 cm^3 = 8.50 ×  10⁻⁶ m³

the coefficient of volume expansion of mercury is 1.80 × 10⁻⁴ / K

The increase in the volume of the mercury =  10⁻³ m³ ×  51.00 × 1.80 × 10⁻⁴

The increase in the volume of the mercury = 9.18*10^{-6} \ m^3

Increase in volume of the glass =  10⁻³ × 51.00 × \beta _{glass}

Now; the mercury overflow = Increase in volume of the mercury - increase in the volume of the flask

the mercury overflow = (9.18*10^{-6}  -  51.00* \beta_{glass}*10^{-3})\ m^3

8.50*10^{-6} = (9.18*10^{-6}  -51.00* \beta_{glass}* 10^{-3} )\ m^3

8.50*10^{-6} - 9.18*10^{-6} = ( -51.00* \beta_{glass}* 10^{-3} )\ m^3

-6.8*10^{-7} =  ( -51.00* \beta_{glass}* 10^{-3} )\ m^3

6.8*10^{-7} =  ( 51.00* \beta_{glass}* 10^{-3} )\ m^3

\dfrac{6.8*10^{-7}}{51.00 * 10^{-3}}=  ( \beta_{glass} )

\mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

Thus; the coefficient of volume expansion of the glass is \mathbf{  ( \beta_{glass} )= 1.333 *10^{-5} / K}

5 0
3 years ago
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