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2 years ago

A car drives to the right. There is a large amount of air resistance, and the

Physics

1 answer:

wariber [46]2 years ago

5 0

Answer:

Depending on what figure you're looking at,

Explanation:

But the answer to some may be Force A

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The current supplied by a battery slowly decreases as the battery runs down. Suppose that the current as a function of time is:

ludmilkaskok [199]

Answer: 8.1 x 10^24

Explanation:

I(t) = (0.6 A) e^(-t/6 hr)

I'll leave out units for neatness: I(t) = 0.6e^(-t/6)

If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).

For neatness let k = 1/(6x3600) = 4.63x10^-5, then:

I(t) = 0.6e^(-kt)

Providing t is in seconds, total charge Q in coulombs is

Q= ∫ I(t).dt evaluated from t=0 to t=∞.

Q = ∫(0.6e^(-kt)

= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.

= -(0.6/k)[e^-∞ - e^-0]

= -0.6/k[0 - 1]

= 0.6/k

= 0.6/(4.63x10^-5)

= 12958 C

Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.

5 0

2 years ago

A steel rod with a length of l = 1.55 m and a cross section of A = 4.45 cm2 is held fixed at the end points of the rod. What is

Blababa [14]

To solve this problem it is necessary to apply the concepts related to thermal stress. Said stress is defined as the amount of deformation caused by the change in temperature, based on the parameters of the coefficient of thermal expansion of the material, Young's module and the Area or area of the area.

$F = AY\alpha \Delta T$

Where

A = Cross-sectional Area

Y = Young's modulus

$\alpha$ = Coefficient of linear expansion for steel

$\Delta T$ = Temperature Raise

Our values are given as,

$A = 4.45cm^2$

$T = 37K$

$\alpha = 1.17*10^{-5}K^{-1}$

$Y = 200*10^9Gpa$

Replacing we have,

$F = (4.45*10^{-4})(200*10^9)(1.17*10^{-5})(37)$

$F = 38526.1N$

Therefore the size of the force developing inside the steel rod when its temperature is raised by 37K is 38526.1N

7 0

2 years ago

14 gauge copper wire has a diameter of 1.6 mm. what length of this wire has a resistance of 4.8ω?

Vladimir79 [104]

The relationship between resistance R and resistivity $\rho$ is
$R= \frac{\rho L}{A}$
where L is the length of the wire and A its cross section.

The radius of the wire is half the diameter:
$r= \frac{d}{2}= \frac{1.6 mm}{2}=0.8 mm=8\cdot 10^{-4} m$
and the cross section is
$A=\pi r^2 = \pi (8\cdot 10^{-4} m)^2=2.01\cdot 10^{-6} m^2$

From the first equation, we can then find the length of the wire when $R=4.8 \Omega$ (copper resistivity: $\rho = 1.724 \cdot 10^{-8} \Omega m$ )
$L= \frac{AR}{\rho}= \frac{(2.01\cdot 10^{-6} m^2)(1.724 \cdot 10^{-8} \Omega m)}{4.8 \Omega}=7.21 \cdot 10^{-15} m$

4 0

2 years ago

The diagram shows two vectors that point west and north. What is the magnitude of the resultant vector? 13 miles 17 miles 60 mil

Kipish [7]

Using the formula A squared plus B squared equals C squared, we can find the solution by substituting 5 for A and 12 for B.

By squaring 5, we get 25, and by squaring 12, we get 144. Adding these, we get 169. The square root of this is 13.

6 0

2 years ago

Read 2 more answers

A point charge Q = -400 nC and two unknown point charges, q1 and q2, are placed as shown. Point charge q1 is located 1.3 meters

nalin [4]

120 nC is the answer

Sorry if I’m wrong

6 0

2 years ago