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3 years ago

13

The passenger-side rear view mirror on a car says, “Objects in the mirror may be closer than they appear”. Assuming the images a

re not inverted, this mirror must be:

1 answer:

Ipatiy [6.2K]3 years ago

4 0

Answer:

the mirror is a convex mirror

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Usain Bolt is the fastest human at 12.4 m/s. If Usain Bolt had a mass of 94kg, what was Usain Bolt’s KE?

Fittoniya [83]

Answer:The fastest recorded human footspeed was recorded between 60 and 80m in Bolt's world record 9.58-second 100m final in Berlin in 2009. Bolt was clocked at 44.72km/h, which is 27.8mph. The Jamaican covered the distance between 60-80m in a time of just 1.61 seconds

Explanation:

pls brainlist too

3 0

3 years ago

Two particles, each with charge Q, and a third charge q, are placed at the vertices of an equilateral triangle as shown. The tot

Gelneren [198K]

Answer:

<em>D. The total force on the particle with charge q is perpendicular to the bottom of the triangle.</em>

Explanation:

The image is shown below.

The force on the particle with charge q due to each charge Q = $\frac{kQq}{r^{2} }$

we designate this force as N

Since the charges form an equilateral triangle, then, the forces due to each particle with charge Q on the particle with charge q act at an angle of 60° below the horizontal x-axis.

Resolving the forces on the particle, we have

for the x-component

$N_{x}$ = N cosine 60° + (-N cosine 60°) = 0

for the y-component

$N_{y}$ = -f sine 60° + (-f sine 60) = -2N sine 60° = -2N(0.866) = -1.732N

The above indicates that there is no resultant force in the x-axis, since it is equal to zero ( $N_{x}$ = 0).

The total force is seen to act only in the y-axis, since it only has a y-component equivalent to 1.732 times the force due to each of the Q particles on q.

<em>The total force on the particle with charge q is therefore perpendicular to the bottom of the triangle.</em>

5 0

4 years ago

You work for an advertising company and have been hired to place a blimp above a football stadium. The angle of elevation from a

iogann1982 [59]

Answer:

a) tangent ; b) 153.88 yds ; c) No , less than or equal to 45° ; D) 45°

Explanation:

Given the following ;

From the triangle sketch :

Base length = 50 yards

Angle of elevation = 72°

a. Which trigonometric ratio would you use to calculate how high the blimp will be above the 50 yard line?

Using trigonometry :

The height of the blimp will be calculated using :

Tangent :

Tan θ = opposite / Adjacent

B.) How high above the ground is the blimp?

Using :

Tan θ = opposite / Adjacent

Θ = 72° ; adjacent = 52, opposite = height(h)

Tan 72° = h / 50

h = 3.0776835 * 50

h = 153.88 yds

C.) In order to be able to read the advertisement on the side of the blimp the highest the blimp can be is 150 feet. Will the fans be able to read the advertisement?

1 yard = 3 Feets

153.88 yards = 3 * 153.88

= 461.65 feets

No, because the height of the blimp is 461.65 Feets which is greater than 150 Feets.

To make viewing possible, the angle of elevation should be:

50 yards is equivalent to (3 * 50) = 150 feets

Max imum Height of blimp = 150 Feets

From pythagoras ;

Tanθ = 150 Feets / 150 Feets

Tanθ = 1

θ = tan^-1(1)

θ = 45°

To make viewing advertisement possible, angle of elevation should not exceed 45°

d.)If height of blimp is 150 Feets, then the exact angle of elevation will be 45°

5 0

3 years ago

Long, long ago, on a planet far, far away, a physics experiment was carried out. First, a 0.210-kg ball with zero net charge was

tigry1 [53]

Answer:

$\Delta V=316167V$

Explanation:

The difference of electric potential between two points is given by the formula $\Delta V=Ed$ , where <em>d</em> is the distance between them and<em> E</em> the electric field in that region, assuming it's constant.

The electric field formula is $E=\frac{F}{q}$ , where <em>F </em>is the force experimented by a charge <em>q </em>placed in it.

Putting this together we have $\Delta V=\frac{Fd}{q}$ , so we need to obtain the electric force the charged ball is experimenting.

On the second drop, the ball takes more time to reach the ground, this means that the electric force is opposite to its weight <em>W</em>, giving a net force $N=W-F$ . On the first drop only <em>W</em> acts, while on the second drop is <em>N</em> that acts.

Using the equation for accelerated motion (departing from rest) $d=\frac{at^2}{2}$ , so we can get the accelerations for each drop (1 and 2) and relate them to the forces by writting:

$a_1=\frac{2d}{t_1^2}$

$a_2=\frac{2d}{t_2^2}$

These relate with the forces by Newton's 2nd Law:

$W=ma_1$

$N=ma_2$

Putting all together:

$N=W-F=ma_1-F=ma_2$

Which means:

$F=ma_1-ma_2=m(a_1-a_2)=m(\frac{2d}{t_1^2}-\frac{2d}{t_2^2})=2md(\frac{1}{t_1^2}-\frac{1}{t_2^2})$

And finally we substitute:

$\Delta V=\frac{Fd}{q}=\frac{2md^2}{q}(\frac{1}{t_1^2}-\frac{1}{t_2^2})$

Which for our values means:

$\Delta V=\frac{2(0.21Kg)(1m)^2}{7.7\times10^{-6}C}(\frac{1}{(0.35s)^2}-\frac{1}{(0.65s)^2})=316167V$

7 0

3 years ago

What is the frequency of light for which the wavelength is 7.1 × 102 nm?

Brilliant_brown [7]

Answer:

Frequency, $f=4.22\times 10^{14}\ Hz$

Explanation:

Given that,

Wavelength of the light,

$\lambda=7.1\times 10^2\ nm=7.1\times 10^2\times 10^{-9}\ m\\\\\lambda=7.1\times 10^{-7}\ m$

We need to find the frequency of light. We know that light is an electromagnetic wave. It moves with the speed of light. So,

$c=f\lambda$

f is the frequency of light

$f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8\ m/s}{7.1\times 10^{-7}\ m}\\\\f=4.22\times 10^{14}\ Hz$

So, the frequency of light is $4.22\times 10^{14}\ Hz$ . Hence, this is the required solution.

5 0

4 years ago