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Tresset [83]
3 years ago
10

A car starts from rest and accelerates with a constant acceleration of 1.00 m/s2 for 3.00 s. The car continues for 5.00 s at con

stant velocity. How far has the car traveled from its starting point?
Physics
1 answer:
barxatty [35]3 years ago
8 0

Answer:

19.5 m

Explanation:

Using the equation of motion,

For the first 3 seconds,

s = ut +1/2at².......................... Equation 1

Where t = time, u = initial velocity, a = acceleration, s = distance

Note: Since the car starts from rest, u = 0 m/s.

Given: a = 1.0 m/s, t = 3 s, u = 0 m/s

Substitute into equation 1

s = 0×3+1/2(3)²(1)

s = 9/2

s = 4.5

In the remaining five seconds, at a constant velocity,

s' = vt............. Equation 2

Where v = velocity, t = time.

Recall,

v = u+at

v = 0+1(3)

v = 3 m/s.

Also, t = 5 s.

Substitute into equation 2

s' = 3(5)

s' = 15 m.

Total distance = 15+4.5

Total distance = 19.5 m.

Hence the car travels 19.5 m

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A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated
Lelu [443]

Incomplete question as we have not told to find what quantity.The complete question is here

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 5.00 cm.calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

Answer:

(a) C=16.7pF

(b) r_{a} =3.749cm

(c) E=2.24*10^{4} N/C

Explanation:

Given data

Q=3.50nC\\V=210V\\r_{b}=5.0cm

For part (a)

The Capacitance given by:

C=\frac{Q}{V}\\ C=\frac{3.50*10^{-9} C}{210V}\\C=1.6666*10^{-11}F\\or\\C=16.7pF

For part (b)

The Capacitance of coordinates is given as

C=\frac{4\pi e}{\frac{1}{r_{a} }-\frac{1}{r_{b} } }\\ So\\{\frac{1}{r_{a} }-\frac{1}{r_{b} } }=\frac{4\pi *8.85*10^{-12} }{1.666*10^{-11}}=6.672m^{-1} \\ \frac{1}{r_{a} }=6.672+(1 /0.05)\\\frac{1}{r_{a} }=26.672\\r_{a} =1/26.672\\r_{a} =0.0375m\\r_{a} =3.749cm

For part (c)

The electric field according to Gauss Law is given by:

EA=\frac{Q}{e}\\ E=\frac{Q}{4\pi er_{a}^{2}  }=\frac{kQ}{r_{a}^{2}}\\  E=\frac{9*10^{9}*3.50*10^{-9} }{(0.0375m)^{2} }\\ E=2.24*10^{4} N/C

7 0
3 years ago
What is the wave speed of a wave that has a frequency of 250 Hz and a wavelength of 0.35 m?
azamat

Answer:

87.5 m/s

Explanation:

The speed of a wave is given by

v=\lambda f

where

v is the wave speed

\lambda is the wavelength

f is the frequency

In this problem, we have

f=250 Hz is the frequency

\lambda=0.35 m is the wavelength

Substituting into the equation, we find

v=(0.35 m)(250 Hz)=87.5 m/s

6 0
3 years ago
A screen is placed 1.20m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.40cm
andrew11 [14]

Answer:

2.8 cm

Explanation:

y_1 = Separation between two first order diffraction minima = 1.4 cm

D = Distance of screen = 1.2 m

m = Order

Fringe width is given by

\beta_1=\dfrac{y_1}{2}\\\Rightarrow \beta_1=\dfrac{1.4}{2}\\\Rightarrow \beta_1=0.7\ cm

Fringe width is also given by

\beta_1=\dfrac{m_1\lambda D}{d}\\\Rightarrow d=\dfrac{m_1\lambda D}{\beta_1}

For second order

\beta_2=\dfrac{m_2\lambda D}{d}\\\Rightarrow \beta_2=\dfrac{m_2\lambda D}{\dfrac{m_1\lambda D}{\beta_1}}\\\Rightarrow \beta_2=\dfrac{m_2}{m_1}\beta_1

Distance between two second order minima is given by

y_2=2\beta_2

\\\Rightarrow y_2=2\dfrac{m_2}{m_1}\beta_1\\\Rightarrow y_2=2\dfrac{2}{1}\times 0.7\\\Rightarrow y_2=2.8\ cm

The distance between the two second order minima is 2.8 cm

8 0
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krek1111 [17]

Answer:

The straight line that is obtained, intercept it on the y-axis and the value of displacement will obtained.

Explanation:

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Answer:

B

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FOR EXAMPLE: OBSERVATIONS AND EXPERIMENTS TO OBTAIN RESULTS.

ANYWAY I LEAVE YOU THE LINK:

https://gscourses.thinkific.com

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