A candle 4.75 cm tall is 40.0 cm to the left of a plane mirror.
a) Where is the image formed by the mirror?
- to the right of the mirror
- to the left of the mirror
b) What is the height of this image?
(Express your answer in centimeters.)

1. The speed with which the ball hits the ground is 17.1 m/s

2. The magnitude of the average force of air resistance exerted on it is 0.77 N

v² = u² + 2gh

v² = 2gh

Take the square root of both side

v = √(2 × 9.8 × 15)

v = 17.1 m/s

We'll begin by calculating the time to reach the ground. This is illustrated below:

h = ½gt²

15 = ½ × 9.8 × t²

15 = 4.9 × t²

Divide both side by 4.9

t² = 15 / 4.9

Take the square root of both side

t = √(15 / 4.9)

t = 1.75 s

Now we can determine the force. This can be obtained as illustrated below:

F = m(v –u) / t

F = 0.149(9 – 0) / 1.75

F = 0.77 N

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Answer:

unfournatletly

Explanation:

i have no clue sorry to waste ur time ill rather not say a answer that will be incorrect.

Answer:

Pluto

Explanation:

The vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

Apply the principle of moment and calculate the tension in the cable;

Clockwise torque = TL sinθ

Anticlockwise torque = ¹/₂WL

TL sinθ  =  ¹/₂WL

T sinθ  =  ¹/₂W

T = (W)/(2 sinθ)

T = (29 x 9.8)/(2 x sin57)

T = 169.43 N

T + F = W

F = W - T

F = (9.8 x 29) - 169.43

F = 114.77 N

Thus, the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

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Answer:

b

Explanation:

the answer is b because if you do the math it equals to b

The orbital speed of an ice cube in the rings of Saturn is determined as  11,237.7 m/s.

The orbital speed of an astronomical body or object is the speed at which it orbits around the center of mass of the most massive body.

The orbital speed of ice cube in the rings of Saturn is calculated as follows;

v = √GM/r

where;

v = √(6.67 x 10⁻¹¹ x 5.68 x 10²⁶)/(3 x 10⁸)

v = 11,237.7 m/s

Thus, the orbital speed of an ice cube in the rings of Saturn is determined as  11,237.7 m/s.

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The kind of equation that can be used to differentiate the kind of separatrix that shows change on motion is

H = 2g/l.

A simple pendulum can be defined as the equipment that displays an oscillatory motion when a mass is tied on a rope and is suspended from it.

The various movements that occur using a simple pendulum is translational ( side to side) or continuous circle (oscillatory motion).

The equation that show that a change from one type of motion to another is H = 2g/l.

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The speed of the trimmer is not affected by the the distribution of trimmer line inside the spool. Option A

The weed trimmer is a device that is used to trim the grasses on a lawn or a field. This device has a rotating shaft that does the actual trimming of the grasses.

The speed of the trimmer is not affected by the the distribution of trimmer line inside the spool thus it does not affect the amount of time required to reach 9,000 rpm speed.

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The speed of the space craft relative to the earth is given as: 0.024c. This is solved using the the equation for time dilation.

Time dilation is the "slowing down" of a clock as determined by an observer in relative motion with regard to that clock under the theory of special relativity.

The formula is given as :

Δt = [Δr]/ √ 1 - (v²/c²)

Thus,

v = c√1 - (Δr/Δt)²

= c √(1 - (3600/3601)²

v = 0.024c

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The range of horizontal projectile motion is 3737.5 m.

A projectile's horizontal velocity is constant (a never changing value), Gravity causes a downward vertical acceleration with a magnitude of 9.8 m/s/s. A projectile's vertical velocity fluctuates by 9.8 m/s per second, A projectile's vertical motion is unrelated to its horizontal motion.  we know that sinθ is maximum at 90°. Therefore horizontal range will maximum at 45°.

To calculate the range of horizontal projectile motion we use;

Δx = vₓ t

where , Δx = Range

             vₓ  = Velocity

              t   = Time

Initial horizontal velocity, vₓ = 650 m/s

Time = 5.75 sec

                Δx = 650 × 5.75

                     = 3737.5 m

Therefore, the range of horizontal projectile motion is 3737.5 m.

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The magnitude of the tension in the string marked A is 52.5N

Generally, the equation for is mathematically given as

Let's take θ be an angle at A

So, tanθ = 3/8

Let's take α be an angle at B (Below X)

tanα = 5/4

Let's take β be an angle at C (Below x)

tanβ = 1/6

First we take the Horizontal Components

74.9cos(9.46°) = Acos(20.6°) + Bcos(51.3°)

By solving the equation, we get

A = 78.9 - 0.668B … (1)

Now, we take the vertical components

74.9sin(9.46°) + Asin(20.6°) = Bsin(51.3°)

By solving the equation, we get

40.07 = 1.015B

B = 39.5N

By substituting the value of B in equation (1)

A = 78.9 - 0.6668× 39.5

A = 52.5N

Hence, the magnitude of the tension in the string marked A is 52.5N

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The tension in the string holding the tassel and the vertical will the tension in the string

Generally, the equation for Tension is  mathematically given as

[tex]TCos\theta = mg[/tex]

Therefore

[tex]TCos6.58^{o} = 19.8*10^{-3}*9.8[/tex]

T = 0.1953 N

b).

Where

[tex]T* sin \theta = ma[/tex]

[tex]0.1953*Sin6.58 \textdegree = 19.8*10^{-3}*a[/tex]

a = 1.13 m/s^2

In conclusion

T* sinФ = ma

2msinФ = ma

2sinФ = a

[tex]sin\theta = \frac{a}{2}[/tex]

[tex]\theta = sin^{-1}\frac{a}{2} \\\\\theta= sin^{-1}\frac{1.13}{2}[/tex]

Ф = 34.4 °

In conclusion, The tension in the string holding the tassel and the vertical will the tension in the string

T = 0.1953 N

Ф = 34.4 °

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Answer:

Explanation:

2.1 x 10^2 - 20J

The image is present at 20cm from the crown glass spherical surface.

To find the answer, we need to know about the lens formula.

=> (1/V)+5=10

=> 1/V= 5

=> V=0.2m = 20cm

Thus, we can conclude that the image is present at 20cm.

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The total current through the battery is 4.77 A.

The current through resistor R1 is 2.4 A.

The current through resistor R2 is 1.2 A

The current through resistor R3 is 0.8 A

The current through resistor R4 is 0.36 A.

The total resistance of the circuit is calculated as follows;

1/Rt = 1/R₁ + 1/R₂ + 1/R₃ + 1/R₄

1/Rt = 1/5 + 1/10 + 1/15 + 1/33

1/Rt = 0.397

Rt = 1/0.397

Rt = 2.518 ohms

I = V/Rt

I = 12/2.518

I = 4.77 A

I₁ = V/R₁

I₁ = 12/5

I₁ = 2.4 A

I₂ = 12/10

I₂ = 1.2 A

I₃ = 12/15

I₃ = 0.8 A

I₄ = 12/33

I₄ = 0.36 A

Thus, the total current through the battery is 4.77 A.

The current through resistor R1 is 2.4 A.

The current through resistor R2 is 1.2 A

The current through resistor R3 is 0.8 A

The current through resistor R4 is 0.36 A.

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Answer:

Average density of the crown: approximately [tex]8\; {\rm g \cdot mL^{-1}}[/tex].

Hence, if this crown contains no empty space, this crown is not made of pure gold.

Explanation:

Let [tex]m(\text{crown})[/tex] and [tex]V(\text{crown})[/tex] denote the mass and volume of this crown. Let [tex]g[/tex] denote the gravitational field strength.

Since this crown is fully immersed in water, the volume of water displaced [tex]V(\text{water, displaced})[/tex] is equal to the volume of this crown:

[tex]V(\text{water, displaced}) = V(\text{crown})[/tex].

The mass of water displaced would be:

[tex]\begin{aligned}m(\text{water, displaced}) &= \rho(\text{water}) \, V(\text{water, displaced}) \\ &= \rho(\text{water}) \, V(\text{crown})\end{aligned}[/tex].

The weight of water displaced would be [tex]m(\text{water, displaced})\, g = \rho(\text{water}) \, V(\text{crown})\, g\end{aligned}[/tex].

The buoyancy force on this crown is equal to the weight of water that this crown displaced:

[tex]F(\text{buoyancy}) = \rho(\text{water}) \, V(\text{crown})\, g[/tex].

The magnitude of this buoyancy force is [tex]7.84\; {\rm N} - 6.86\; {\rm N} = 0.98\; {\rm N}[/tex]. Rearrange the equation for buoyancy to find [tex]V(\text{crown})[/tex]:

[tex]\begin{aligned} V(\text{crown}) &= \frac{F(\text{buoyancy}) }{\rho(\text{water}) \, g}\end{aligned}[/tex].

Since the weight of this crown is [tex]\text{weight}(\text{crown}) = m(\text{crown})\, g[/tex], the mass of this crown would be [tex]m(\text{crown})= \text{weight}(\text{crown}) / g[/tex].

The average density of this crown would be:

[tex]\begin{aligned}\rho(\text{crown}) &= \frac{m(\text{crown})}{V(\text{crown})} \\ &= \frac{\text{weight}(\text{crown}) / g}{F(\text{buoyancy}) / (\rho(\text{water})\, g)} \\ &= \frac{\text{weight}(\text{crown})}{F(\text{buoyancy})}\, \rho(\text{water}) \\ &= \frac{7.84\; {\rm N}}{0.98\; {\rm N}}\times 1.000 \; {\rm g\cdot mL^{-1}} \\ &= 8.0\; {\rm g \cdot mL^{-1}}\end{aligned}[/tex].

The density of pure gold is significantly higher than [tex]8.0\; {\rm g\cdot mL^{-1}}[/tex]. Hence, if this crown contains no empty space (i.e., no air bubble within the crown), the crown would not be made of pure gold.

(a) The minimum force F he must exert to get the block moving is 38.9 N.

(b) The acceleration of the block is 0.79 m/s².

The minimum force F he must exert to get the block moving is calculated as follows;

Fcosθ = μ(s)Fₙ

Fcosθ = μ(s)mg

where;

F = [0.1(36)(9.8)] / [(cos(25)]

F = 38.9 N

F(net) = 38.9 - (0.03 x 36 x 9.8) = 28.32

a = F(net)/m

a = 28.32/36

a = 0.79 m/s²

Thus, the minimum force F he must exert to get the block moving is 38.9 N.

The acceleration of the block is 0.79 m/s².

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The point where m3 experiences a zero net gravitational force due to M1 and m2 is 57.42 m.

m1<------(x)------> m3 <-----------(94.8 m - x)-------->m2

a point, x, where m3 experiences a zero net gravitational force due to M1 and m2;

Force on m3 due to m1 = Force on m3 due to m2

Gm1m3/d² = Gm2m3/r²

m1/d² = m2/r²

where;

m1/(x²) = m2/(94.8 - x)²

m1(94.8 - x)² = m2x²

(94.8 - x)² = (m2/m1)x²

(94.8 - x)² = (10.6/25)x²

(94.8 - x)² = 0.424x²

(94.8 - x)² = (0.651)²x²

94.8 - x = 0.651x

94.8 = 1.651x

x = 94.8/1.651

x = 57.42 m

Thus, the point where m3 experiences a zero net gravitational force due to M1 and m2 is 57.42 m.

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Answer:

0.09kg of air

Explanation:

The dimensions of the room are given

change the height to meters by dividing it by thousand.

For the volume multiplying the length,width and height (all should be in the same unit most suitable being meters).

Volume refers to the amount of space inside a box or a object.

The amount of air is equal to the volume.

Answer:

90 m^3

Explanation:

Volume of the room:

    6 m * 5 m * 3 m         =  90 m^3   <=====( I changed 3mm to 3 m)

if  3mm is not a typo mistake

 volume becomes     ( 3 mm = .003 m)

      6 m * 5 m * .003 m   = .09 m^3   ( though unlikely )

The average force exerted by the ball on the glove is 480 N.

The average force exerted on the glove by the ball is equal in magnitude to the force on the ball.

mass = 145 g = 0.145 kg

Acceleration of the baseball, a = (v² - u²)/2s

where:

v is final velocity = 0

u is initial velocity = 39 m/s

s is distance = -23 cm = 0.23 m

a = (0 - 39²)/2(-0.23)

a = 3306.52 m/s²

Force = 0.145 * 3306.52

Force = 479.4 N

Average force = 480 N

In conclusion, force is derived from the product of mass and acceleration.

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The maximum acceleration the truck can have so that the refrigerator does not tip over is 1.86m/s².

Acceleration is the change in velocity with time.

The maximum acceleration is obtained by taking moments about the tipping point of rotation.

F₂ * 1.58 m = F₁ * 0.67 m

where

F₂ is tipping force = mass * acceleration, a

F₁ is weight = mass * acceleration due to gravity, g

The weight acts at a distance half the width of the refrigerator = 30 cm or 0.3 m

Height of refrigerator is 158 cm 0r 1.58 cm

m * a * 1.58 = m * 9.81 * 0.30

a = 1.86 m/s²

The maximum acceleration the truck can have so that the refrigerator does not tip over is 1.86m/s².

In conclusion, if the maximum acceleration is exceeded, the refrigerator will tip over.

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The tangential speed of the wheel is determined as 4.786 m/s.

The tangential speed of the wheel is calculated as follows;

v = ωr

where;

v = (2.17 x 2π rad)/s x 0.351 m

v = 4.786 m/s

Thus, the tangential speed of the wheel is determined as 4.786 m/s.

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The acceleration due to gravity is 3 m/s^2 and the work done is -4.3 J.

Now we must use the formula;

h = ut + 1/2gt^2

Since it was dropped from a height u = 0 m/s

h = height

u = initial velocity

g = acceleration due to gravity

t = time

h = 1/2gt^2

g = 2h/t^2

g = 2 * 2.85 /(1.38)^2

g = 5.7/1.9

g = 3 m/s^2

The work that must be done is against gravity hence;

W = -(mgh)

W = - (0.5 kg *  3 m/s^2 * 2.85 m)

W = - 4.3 J

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The direction of magnetic field is south-east and the magnitude is

[tex]23.66[/tex] × [tex]10^{3} N/C[/tex].

Here, the magnitude of CD and BC will be cancelled, as they both are in the opposite direction and equal to each other.
the magnitude, towards the diagonal AC will result in CP, which is the direction of the electric field.

magnitude of electric field can be defined as :- The force per charge on the test charge is a straight forward way to define the size of the electric field.

To find the magnitude of the electric field use the formula

[tex]E = kq/ r^{2}[/tex]

inserting the values,

[tex]E = 9. 10^{9}[/tex] × [tex]4.05[/tex] × [tex]10^{-6} / 1.1 \sqrt{2}[/tex]

[tex]E= 36.45[/tex] × [tex]10^{3}[/tex] / [tex]1.54[/tex]

[tex]E = 23.66[/tex] × [tex]10^{3}[/tex] N/C

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In my opinion, I think that the input data that are required to find the least energy path are:

This refers to the use of a diagram to represent the features of a place that shows its physical landforms to help in navigation.

Hence, we can see that when using maps like gmaps, it is important to consider both elevation and distance to be able to find the path that uses the least energy.

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The mass of the planet Mar, given the data from the question is 6.45×10²³ Kg

g = GM / r²

Cross multiply

GM = gr²

Divide both sides by G

M = gr² / G

M = (3.724 × 3400000²) / 6.67×10¯¹¹

M = 6.45×10²³ Kg

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Answer:

80GB= 80000000000 bytes

256GB= 274877906944 bytes

Explanation:80GB= 80000000000 bytes

256GB= 274877906944 bytes

The Answer is Option C. Friction.

Explanation:

The friction force acts in the forward direction on the rear wheel and it acts in the backward direction on the front wheel. The magnitude of friction force on the rear wheel can be more, equal or less than that on the front wheel.

Answer:

[tex]\alpha=arccos[\frac{(a^2+b^2)(n-1)}{2ab(n+1)}].[/tex]

Explanation:

1) for A+B: a²+b²-2abcos(π-α);

2) for A-B: a²+b²-2abcos(α);

3) according to the condition (A+B):(A-B)=n, then

[tex]n=\frac{a^2+b^2-2abcos(\pi-a)}{a^2+b^2-2abcos(a)}; \ = > \ n=\frac{a^2+b^2+2abcos(a)}{a^2+b^2-2abcos(a)}; \ = > \ cos(\alpha)=\frac{(a^2+b^2)(n-1)}{2ab(n-1)}.[/tex]

Relative to the positive horizontal axis, rope 1 makes an angle of 90 + 20 = 110 degrees, while rope 2 makes an angle of 90 - 30 = 60 degrees.

By Newton's second law,

[tex]R_1 \cos(110^\circ) + R_2 \cos(60^\circ) = 0[/tex]

where [tex]R_1,R_2[/tex] are the magnitudes of the tensions in ropes 1 and 2, respectively;

[tex]R_1 \sin(110^\circ) + R_2 \sin(60^\circ) - mg = 0[/tex]

where [tex]m=1300\,\rm kg[/tex] and [tex]g=9.8\frac{\rm m}{\mathrm s^2}[/tex].

Eliminating [tex]R_2[/tex], we have

[tex]\sin(60^\circ) \bigg(R_1 \cos(110^\circ) + R_2 \cos(60^\circ)\bigg) - \cos(60^\circ) \bigg(R_1 \sin(110^\circ) + R_2 \sin(60^\circ)\bigg) = 0\sin(60^\circ) - mg\cos(60^\circ)[/tex]

[tex]R_1 \bigg(\sin(60^\circ) \cos(110^\circ) - \cos(60^\circ) \sin(110^\circ)\bigg) = -\dfrac{mg}2[/tex]

[tex]R_1 \sin(60^\circ - 110^\circ) = -\dfrac{mg}2[/tex]

[tex]-R_1 \sin(50^\circ) = -\dfrac{mg}2[/tex]

[tex]R_1 = \dfrac{mg}{2\sin(50^\circ)} \approx \boxed{8300\,\rm N}[/tex]

Solve for [tex]R_2[/tex].

[tex]\dfrac{mg\cos(110^\circ)}{2\sin(50^\circ)} + R_2 \cos(60^\circ) = 0[/tex]

[tex]\dfrac{R_2}2 = -mg\cot(110^\circ)[/tex]

[tex]R_2 = -2mg\cot(110^\circ) \approx \boxed{9300\,\rm N}[/tex]