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adoni [48]
3 years ago
15

Please answer the question number 2 in this pic in detail

Physics
1 answer:
soldier1979 [14.2K]3 years ago
3 0
It would be C ............
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What force in Newton is required to accelerate a car starting from rest to 20 m/s in 15 seconds if the mass of the car is 2500 k
Lyrx [107]

We will solve this question using the second law of motion which states that force is directly equal to the product of mass and acceleration.

\sf \: F=ma

Where,

  • F is force
  • m is mass
  • a is acceleration

In our case,

  • F = ?
  • m = 2500 kg
  • a = 20m/s

\tt \: F_{net}  = 2500 \times 20 \\   \tt= 50000

<em>Thus, The force of 50000 Newton is required to accelerate a car of 2500 kg...~</em>

3 0
2 years ago
Read 2 more answers
What is the difference between a wedge and a screw?
Anvisha [2.4K]
<span>A wedge is a double sided screw basically. Therefore, the difference is the number of sides.</span>
7 0
3 years ago
Using equations, determine the temperature, pressure and density of the air for a aircraft flying at 19.5 km. Is this aircraft s
Viefleur [7K]

Answer:

a) - 72.5°c

b) pressure = 3625.13 Pa

c) density =  0.063 kg/m^3

d) it is a subsonic aircraft

Explanation:

a) Determine Temperature

Temperature at 19.5 km ( 19500 m )

T = -131 + ( 0.003 * altitude in meters )

  =  -131 + ( 0.003 * 19500 ) = - 72.5°c

b) Determine pressure and density at 19.5 km altitude

Given :

Po (atmospheric pressure at sea level )  = 101kpa

R ( gas constant of air ) = 0.287 KJ/Kgk

T = -72.5°c ≈ 200.5 k

pressure = 3625.13 Pa

hence density = 0.063 kg/m^3

attached below is the remaining part of the solution

C) determine if the aircraft is subsonic or super sonic

Velocity ( v ) = \sqrt{CRT}  =  \sqrt{1.4*287*200.5 } = 283.8 m/s

hence it is a subsonic aircraft

4 0
3 years ago
A high-pass filter consists of a 1.66 μF capacitor in series with a 80.0 Ω resistor. The circuit is driven by an AC source with
Julli [10]

Explanation:

Given that,

Capacitor C=1.66\ \mu F

Resistor R=80.0\ \Omega

Peak voltage = 5.10 V

(A). We need to calculate the crossover frequency

Using formula of frequency

f_{c}=\dfrac{1}{2\pi R C}

Where, R = resistor

C = capacitor

Put the value into the formula

f_{c}=\dfrac{1}{2\pi\times80.0\times1.66\times10^{-6}}

f_{c}=1198.45\ Hz

(B). We need to calculate the V_{R} when f = \dfrac{1}{2f_{c}}

Using formula of  V_{R}

V_{R}=V_{0}(\dfrac{R}{\sqrt{R^2+(\dfrac{1}{2\pi fC})^2}})

Put the value into the formula

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times\dfrac{1}{2}\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=2.280\ Volt

(C). We need to calculate the V_{R} when f = f_{c}

Using formula of  V_{R}

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=3.606\ Volt

(D). We need to calculate the V_{R} when f = 2f_{c}

Using formula of  V_{R}

V_{R}=5.10\times(\dfrac{80.0}{\sqrt{(80.0)^2+(\dfrac{1}{2\pi\times2\times1198.45\times1.66\times10^{-6}})^2}})

V_{R}=4.561\ Volt

Hence, This is the required solution.

8 0
3 years ago
3. Suppose that you have an electrically charged stick. If you divide the stick in half, each half will have half the original c
FrozenT [24]

Answer:

No, you can't keep on dividing the charge forever.

Explanation:

No, you can't keep on dividing the charge in that manner forever because the total charge of the stick is an integer multiples of individual units known as an elementary charge, <em>which is the electron (e) charge (e = 1.602x10⁻¹⁹C)</em>.

Therefore the limit of the division of the original charge will be the electron charge since it is the smallest charge that can exist freely.  

I hope it helps you!  

8 0
3 years ago
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