A small remote controlled car with mass 1.60 kg moves at a constant speed of12.0 m/s in a vertical circle with a radius of 5.0 m.What is the magnitude ofthe normal force exerted on the car by the walls of the circle at pointA?

Answer:

1 mm    PLS mark brainliest

Explanation:

Microwaves are electromagnetic waves with wavelengths longer than those of terahertz (THz) wavelengths, but relatively short for radio waves. Microwaves have wavelengths approximately in the range of 30 cm (frequency = 1 GHz) to 1 mm (300 GHz).

Final answer:

The rocket's top speed is 46 m/s.

Explanation:

The rocket's top speed can be found using the equation v = vo + at, where v is the final velocity, vo is the initial velocity, a is the acceleration, and t is the time. In this case, the initial velocity (vo) is 0 m/s since the rocket starts at rest, the acceleration (a) is 10 m/s², and the time (t) is 4.6 s.

Plugging these values into the equation, we get:

v = 0 m/s + (10 m/s²)(4.6 s) = 46 m/s.

Therefore, the rocket's top speed is 46 m/s.

Answer:

The deceleration is [tex]6ms^{-2}[/tex]

Explanation:

Acceleration is change in velocity with respect to time.

[tex]a = \frac{\Delta V}{\Delta t}\\a = \frac{24-6}{3} \\a = 6[/tex]

Answer:    

Compared to windshield the airbag exerts much lesser force  

Explanation:

Impulse is defined as change in momentum of the object when it is acted upon by a force during interval of time

Impulse = Impulsive force *time

I = F*Δt

If the object should be bought to rest from certain velocity there should be change in momentum. If the duration in which the momentum is increased then there would be less force applied and hence less damage.

Airbags are used to reduce the force experience by the people when they are met with accident by extending the time required to stop the momentum.

During the collision, the passenger is carried towards the windshield and if they are stopped by collision with wind shield the force will be larger and more damage.But if they are hit with airbag then the force will be less due to increased time.

The change is momentum will be the same with or without momentum but its the time that decides the impact of force.By making it longer the force become less.

Thus compared to the windshield the airbag exerts much lesser force.

Answer:

That's DNA

Explanation:

DNA(Deoxyribonucleic acid) is the building blocks of life, the Legos for genetically complex life if you will. All living things have DNA, it is the instruction that that organism must follow to specify what exact traits it'll have, For example, I have Blue eyes because Blue eyes is apart of my DNA. The DNA told my body that I was gonna have blue eyes before I was born.

When a ray is emitted from the tip of an object towards the focal point of a mirror, it follows the law of reflection. The location and orientation of the reflected ray depend on the shape of the mirror. Additional rays can be traced from the base of the object to locate the extended image.

When a ray is emitted from the tip of an object towards the focal point of a mirror, it follows the law of reflection. For a concave mirror, the reflected ray passes through the focal point, while for a convex mirror, the reflected ray extends backward through the focal point. The location and orientation of the reflected ray depend on the shape of the mirror. To locate the extended image, additional rays can be traced from the base of the object along the optical axis. All four principal rays run parallel to the optical axis, reflect from the mirror, and then run back along the optical axis. The image of the base of the object is located directly above the image of the tip, as the mirror is symmetrical from top to bottom.

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Ray tracing involves drawing rays from an object through points of interest like the focal point of a mirror and then determining where the reflected rays intersect to locate the image. It is a crucial method in optics to determine the properties of images formed by mirrors and lenses.

To trace the path of a ray emitted from the tip of an object toward the focal point of a mirror, as well as the reflected ray, one must understand the basic principles of ray tracing with mirrors. With a concave mirror, an incident ray that travels parallel to the optical axis will be reflected through the focal point on the same side of the mirror. Conversely, with a convex mirror, a ray that travels parallel to the optical axis will reflect as if it originates from the focal point behind the mirror, forming a virtual focus. To construct the reflected ray, you draw the ray until it reaches the mirror’s surface and then redirect it according to the mirror type's ray tracing rules.

Furthermore, the intersection of the reflected rays (real or virtual) determines the location of the image. If the rays meet in real space, the image is real; if they only appear to intersect upon extension in virtual space, the image is virtual.

To achieve a complete picture of how an image is formed, one must also trace rays from another point on the object. For instance, tracing rays from the base of the object can help determine the orientation of the image. In the case of rays that are collinear with the optical axis, the image will maintain the object's vertical orientation. By tracing at least two different rays following the simple ray tracing rules, one can locate the image formed by mirrors and lenses.

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Answer:

They would find [tex]3.14\times 10^{10}[/tex]galaxies.

Explanation:

Given that,

The number of density of galaxies in the universe is 3×10⁻⁶⁸ galaxies /m³.

Assuming that, the astronomers are observing at the center of sphere.

So, they can observe the sphere of space whose radius 10¹⁰ light years.

1 light year = 10¹⁶ meters

10¹⁰ light years =10¹⁰ .10¹⁶ meters

                         =10²⁶meters

The volume of the space is

=[tex]\frac43 \pi r^3[/tex]

[tex]=\frac43 \pi (10^{26})^3[/tex] m³

[tex]=\frac43 \pi 10^{78}[/tex] m³.

The number of galaxies

= Volume of the space × density

[tex]=(3\times 10^{-68}\ galaxies /m^3)\times(\frac43 \pi . 10^{78}\ m^3)[/tex]

[tex]=10^{10}\pi[/tex] galaxies

= [tex]3.14\times 10^{10}[/tex]galaxies

They would find [tex]3.14\times 10^{10}[/tex]galaxies.

If astronomers could observe all galaxies out to a distance of 10^10 light-years, they would find approximately 10^100 galaxies.

To find the number of galaxies that astronomers would find if they could observe all galaxies out to a distance of 1010 light-years, we can use the average number density of galaxies in the universe. The number density is given as 3 × 10-68 galaxies/m3. We can convert the distance to meters by multiplying by 1016 (since there are 1016 meters in 1 light-year).

Next, we can calculate the volume of space that astronomers would be observing. The volume can be found by multiplying the distance cubed (in meters) by 4/3π. The number of galaxies can then be calculated by multiplying the volume by the number density.

Substituting the given values into the equation, we have:

Volume = (4/3) × π × (1016)3 m3

Number of galaxies = Number density × Volume

After calculating the volume and multiplying it by the number density, we find that astronomers would find approximately 10100 galaxies if they could observe all galaxies out to a distance of 1010 light-years.

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1) 7.5 N

2) 10 N

Explanation:

1)

We can solve this first part of the problem by using Newton's second law of motion, which states that the net force on an object is equal to the product between its mass and its acceleration:

[tex]F=ma[/tex]

where

F is the net force

m is the mass of the object

a is its acceleration

In this problem we have:

m = 1.5 kg is the mass of the ball

[tex]a=5 m/s^2[/tex] is the acceleration

So, the kick force on it was:

[tex]F=(1.5)(5)=7.5 N[/tex]

2)

In this case, the mass of the ball has increased to

m' = 2 kg

We can also solve this part by using again Newton's second law of motion:

[tex]F'=m'a[/tex]

where

F' is the new kick force

m' = 2 kg is the new mass of the ball

a is the acceleration

The acceleration is the same as before,

[tex]a=5 m/s^2[/tex]

Therefore, the new kick force is:

[tex]F'=(2)(5)=10 N[/tex]

Answer:

a) Pi,c = 1066 kgm/s

b) Pi,b = 0.0075 kgm/s  

c) ΔV = - 0.0007 m/s

d) ΔV = - 0.0008 m/s

Explanation:

Given:-

- The mass of the bicycle, mc = 12 kg

- The mass of passenger, mp = 70 kg

- The mass of the bug, mb = 5.0 g

- The initial speed of the bicycle, vpi = 13 m/s

- The initial speed of the bug, vbi = 1.5 m/s

Find:-

a.What is the initial momentum of you plus your bicycle?

b.What is the initial momentum of the bug?

c.What is your change in velocity due to the collision the bug?

d.What would the change in velocity have been if the bug were traveling in the opposite direction?

Solution:-

- First we will set our one dimensional coordinate system, taking right to be positive in the direction of bicycle.

- The initial linear momentum (Pi,c) of the passenger and the bicycle would be:

                       Pi,c = vpi* ( mc + mp)

                       Pi,c = 13* ( 12+ 70 )

                       Pi,c = 1066 kgm/s  

- The initial linear momentum (Pi,b) of the bug would be:

                       Pi,b = vbi*mb

                       Pi,b = 0.005*1.5

                       Pi,b = 0.0075 kgm/s  

- We will consider the bicycle, the passenger and the bug as a system in isolation on which no external unbalanced forces are acting. This validates the use of linear conservation of momentum.

- The bicycle, passenger and bug all travel in the (+x) direction after the bug splatters on the helmet.

                       Pi = Pf

                       Pi,c + Pi,b = V*(mb + mc + mp)

Where,    V : The velocity of the (bicycle, passenger and bug) after collision.

                      1066 + 0.0075 = V*( 0.005 + 12 + 70 )

                      V = 1066.0075 / 82.005

                      V = 12.9993 m/s

- The change in velocity is Δv = 13 - 12.9993 =  - 0.00070 m/s      

- If the bug travels in the opposite direction then the sign of the initial momentum of the bug changes from (+) to (-).

- We will apply the linear conservation of momentum similarly.

                      Pi = Pf

                      Pi,c + Pi,b = V*(mb + mc + mp)        

                      1066 - 0.0075 = V*( 0.005 + 12 + 70 )

                      V = 1065.9925 / 82.005

                      V = 12.99911 m/s

- The change in velocity is Δv = 13 - 12.99911 =  -0.00088 m/s      

Answer:

a. The initial momentum of you and your bicycle is 1066 kgm/s.

b. The initial momentum of the bug is 0.0075 kgm/s.

c. The change in velocity due to the collision with the bug is -0.0008 m/s.

d. If the bug were travelling in the opposite direction, the change in velocity due to the collision would have been -0.0009 m/s.

Explanation:

The initial momentum of you and your bicycle can be easily calculated using the definition of momentum:

[tex]p=mv\\\\p=(m_{you}+m_{bicycle})v\\\\p=(70kg+12kg)(13m/s)\\\\p=1066kgm/s[/tex]

So the initial momentum of you plus your bicycle is 1066 kgm/s (a).

The initial momentum of the bug can be obtained in the same way:

[tex]p=mv\\\\p=(0.005kg)(1.5m/s)\\\\p=0.0075kgm/s[/tex]

Then the initial momentum of the bug is 0.0075 kgm/s (b).

Now, since the mass of the bug is much less than your mass, we can think of this as a perfectly inelastic collision. This means that, after the collision, the velocity of you, the bicycle and the bug is the same. From the conservation of linear momentum, we have:

[tex]p_0=p_f\\\\(m_{you}+m_{bicycle})v_{you}+m_{bug}v_{bug}=(m_{you}+m_{bicycle}+m_{bug})v_f\\\\v_f=\frac{(m_{you}+m_{bicycle})v_{you}+m_{bug}v_{bug}}{m_{you}+m_{bicycle}+m_{bug}}\\\\v_f=\frac{(70kg+12kg)(13m/s)+(0.005kg)(1.5m/s)}{70kg+12kg+0.005kg}\\ \\v_f=12.9992m/s[/tex]

As your initial velocity was 13m/s, the change in velocity is of -0.0008 m/s (c).

If the bug were travelling in the opposite direction, its initial velocity would have been negative. So:

[tex]v_f=\frac{(70kg+12kg)(13m/s)-(0.005kg)(1.5m/s)}{70kg+12kg+0.005kg}\\ \\v_f= 12.9991m/s[/tex]

So, in this case the change in velocity is of -0.0009 m/s (d).

Note that the bug is so small that the change in velocity is negligible in most cases. That's why we don't notice when we hit a bug when riding bicycle.

Given that,

Horizontal velocity of the object, v = 20 m/s

Height of the cliff, h = 125 m

We need to find the time that it takes the object to fall to the ground from the cliff is most nearly. It can be calculated using second equation of motion. Let us consider that the initial speed of the object is 0. So,

[tex]h=ut+\dfrac{1}{2}at^2[/tex]

Here, a = g and u = 0

[tex]h=\dfrac{1}{2}at^2\\\\t=\sqrt{\dfrac{2h}{g}} \\\\t=\sqrt{\dfrac{2\times 125}{10}} \\\\t=5\ s[/tex]

So, the object will take 5 seconds to fall to the ground from the cliff.

The time taken for the object to fall to the ground is 5 s

From the question given above, the following data were obtained:

The time taken for the object to fall to the ground can be obtained as follow:

[tex]t = \sqrt{ \frac{2h}{g}} \\ \\ t = \sqrt{ \frac{2 \times 125}{10}} \\ \\ t = 5 \: s \\ \\ [/tex]

Therefore, the time taken for the object to get to the ground is 5 s

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Answer:

(a) 1058.4 J

(b) -10584 J

Explanation:

Parameters given:

Mass of astronaut, m = 72 kg

Distance moved by astronaut, d = 15 m

(a) WORK DONE BY FORCE FROM THE HELICOPTER

Work done is given as the product of Force applied to a body and the distance moved by the body:

W = F * d

The force from the helicopter is given as:

F = m * a

where a = acceleration of the astronaut due to the helicopter

Therefore, the work done is given as:

W = m * a * d

W = 72 * g/10 * 15

W = [tex]\frac{72 * 9.8 * 15}{10}[/tex]

W = 1058.4 J

(b) WORK DONE BY FORCE OF GRAVITY

W = F * d

The force of gravity is given as:

F = -m * g

where g = acceleration due to gravity

The negative sign is due to the fact that the astronaut moves in an opposite direction (upwards) to the force of gravity (Gravity acts downwards)

Therefore, the work done is given as:

W = -m * g * d

W = -72 * 9.8 * 15

W = -10584 J

Answer: B (Longitudinal waves travel through solids, liquids, and gases, Transverse waves only travel through  solids)

Explanation: A longitudinal wave alternately compresses the medium and stretches it out. Solids, liquids and gasses all push back when they are compressed— so they are all able to store energy this way and thus transmit the wave. But, Transverse waves can only go through solids because they have enough shear strength, but liquids and gases don't. ( Transverse waves are the transfer of energy in a motion that is perpendicular to the direction the wave is traveling. Only solids are able to switch it's motion to travel through the wave.)

Answer:

b

Explanation:

Answer:

The statement is not true always.

Explanation:

Given:

Star A appears brighter than Star B.

The apparent brightness of a star depends on two factors:

i) the distance of the star from the earth

ii) the limunosity of the star

If two stars are at same distance from the earth and one has more luminosity than other then ir appears to be more luminous. Also if two stars are at a different distance from the earth, then the star having more luminosity appears to be more luminous.

Thus, the statement is not true.

Answer:

Resistance of the circuit is 820 Ω

Explanation:

Given:

Two galvanometer resistance are given along with its voltages.

Let the resistance is "R" and the values of voltages be 'V' and 'V1' along with 'G' and 'G1'.

⇒ [tex]V=20\ \Omega,\ V_1=30\ \Omega[/tex]

⇒ [tex]G=1680\ \Omega,\ G_1=2930\ \Omega[/tex]

Concept to be used:

Conversion of galvanometer into voltmeter.

Let [tex]G[/tex] be the resistance of the galvanometer and [tex]I_g[/tex] the maximum deflection in the galvanometer.

To measure maximum voltage resistance [tex]R[/tex] is connected in series .

So,

⇒ [tex]V=I_g(R+G)[/tex]

We have to find the value of [tex]R[/tex] we know that in series circuit current are same.

For [tex]G=1680[/tex]                                    For [tex]G_1=2930[/tex]

⇒ [tex]I_g=\frac{V}{R+G}[/tex]   equation (i)                ⇒ [tex]I_g=\frac{V_1}{R+G_1}[/tex] equation (ii)

Equating both the above equations:

⇒ [tex]\frac{V}{R+G} = \frac{V_1}{R+G_1}[/tex]

⇒ [tex]V(R+ G_1) = V_1 (R+G)[/tex]

⇒ [tex]VR+VG_1 = V_1R+V_1G[/tex]

⇒ [tex]VR-V_1R = V_1G-VG_1[/tex]

⇒ [tex]R(V-V_1) = V_1G-VG_1[/tex]

⇒ [tex]R =\frac{V_1G-VG_1}{(V-V_1)}[/tex]

⇒ Plugging the values.

⇒ [tex]R =\frac{(30\times 1680) - (20\times 2930)}{(20-30)}[/tex]

⇒ [tex]R =\frac{(50400 - 58600)}{(-10)}[/tex]

⇒ [tex]R=\frac{-8200}{-10}[/tex]

⇒ [tex]R=820\ \Omega[/tex]

The coil resistance of the circuit is 820 Ω .

Answer:

I think that the liquids molecules are slowing down. Hope this helps!

Explanation:

Please vote me Brainliest

Answer: 128J

Explanation:

[tex]Formula: E_k=\frac{1}{2}mv^2[/tex]

[tex]E_k=\frac{1}{2}(16kg)(4m/s)^2 \\E_k=\frac{1}{2}(16kg)(16m^2/s^2)\\ E_k=\frac{1}{2}(256kg*m^2/s^2)\\ E_k=128kg*m^2/s^2\\or\\E_k=128J[/tex]

Answer:

That is correct

Explanation:

128 J

Answer:

Shift towards the normal

Explanation:

Refraction is defined as the change in direction of light rays when passing through from a medium to another.

The ray can either pass through a less dense medium to a denser medium or from a denser medium to a less dense medium. The light ray bends towards or away from the normal ray(ray perpendicular to the plane) depending whether the travels from less dense to denser or otherwise.

Note that if a ray travels from less dense medium which have a low refractive index like air to a more dense medium like water which have a higher refractive index than air, the refracted ray tends to bend towards the normal, otherwise they bend away from the normal.

Answer:

Shift towards the normal

Explanation:

Explanation:

Power of electric kettle, P = 1 kW

Voltage, V = 220 V

(a) Electric power is given by the formula as follows :

[tex]P=\dfrac{V^2}{R}[/tex]

R is resistance

[tex]R=\dfrac{V^2}{P}\\\\R=\dfrac{(220)^2}{10^3}\\\\R=48.4\ \Omega[/tex]

(b) When connected to a 220 V supply, it takes 3 minutes for the water in the kettle to reach boiling point.

Energy supplied is given by :

[tex]E=P\times t[/tex]

P is power, [tex]P=\dfrac{V^2}{R}[/tex]

[tex]E=\dfrac{V^2}{R}t\\\\E=\dfrac{(220)^2}{48.4}\times 180\\\\E=180000\ J\\\\E=180\ kJ[/tex]

The working resistance of the kettle is 48.35 ohms. The amount of energy supplied is 180,000 J.

To calculate the working resistance of the kettle, we can use Ohm's law, which states that resistance (R) is equal to the voltage (V) divided by the current (I).

To calculate the amount of energy supplied, we need to use the formula E = P * t, where E is energy, P is power, and t is time.

Therefore, The working resistance of the kettle is 48.35 ohms. The amount of energy supplied is 180,000 J.

Answer:

- 58 cm

Explanation:

refractive index, n = 1.6

radius of curvature of left face, R1 = - 65 cm

Radius of curvature of the right face, R2 = 75 cm

Use the lens maker's formula

[tex]\frac{1}{f}=\left ( n-1 \right )\times \left ( \frac{1}{R_{1}}-\frac{1}{R_{2}} \right )[/tex]

[tex]\frac{1}{f}=\left ( 1.6-1 \right )\times \left ( -\frac{1}{65}-\frac{1}{75} \right )[/tex]

[tex]\frac{1}{f}=\left ( 0.6 \right )\times \left ( \frac{-75-65}{75\times 65}\right )[/tex]

f = - 58 cm

Thus, the focal length of the lens is - 58 cm.

Answer:

a) v=-6m/s

b) negative direction

c) 6m/s

d) decreasing

e) for t=2s

f) Yes

Explanation:

The particle position is given by:

[tex]x=4-12t+3t^2[/tex]

a) the velocity of the particle is given by the derivative of x in time:

[tex]v=\frac{dx}{dt}=-12+6t[/tex]

and for t=1s you have:

[tex]v=\frac{dx}{dt}=-12+6(1)^2=-6\frac{m}{s}[/tex]

b) for t=1s you can notice that the particle is moving in the negative x direction.

c) The speed can be computed by using the formula:

[tex]|v|=\sqrt{(-12+6t)^2}=\sqrt{(-12+6)^2}=6\frac{m}{s}[/tex]

d) Due to the negative value of the velocity in a) you can conclude that the speed is decreasing.

e) There is a time in which the velocity is zero. You can conclude that because if t=2 in the formula for v in a), v=0

[tex]v=0=-12+6(t)\\\\t=\frac{12}{6}=2[/tex]

f) after t=3s the particle will move in the negative direction, this because it is clear that 4+3t^2 does not exceed -12t.

Answer:

a)  v (1)  = -6 m/s

b) negative x-direction

c) s ( 1 ) = 6 m/s

d) The speed decreases at t increases from 0 to 2 seconds.

e) At t = 2 s, the velocity is 0

f) No

Explanation:

Given:-

- The position function of the particle:

                     x (t) = 4 - 12t + 3t^2

Find:-

what is its velocity at t = 1 s? (b) Is it moving in the positive or negative direction of x just then? (c) What is its speed just then? (d) Is the speed increasing or decreasing just then? (Try answering the next two questions without further calculation.) (e) Is there ever an instant when the velocity is zero? If so, give the time t; if not, answer no. (f) Is there a time after t = 3 s when the particle is moving in the negative direction of x? If so, give the time t; if not, answer no.

Solution:-

- The velocity function of the particle v(t) can be determined from the following definition:

                              v (t) = d x(t) / dt

                              v (t) = -12 + 6t

- Evaluate the velocity at time t = 1 s:

                              v (1) = -12 + 6(1)

                              v (1)  = -6 m/s

- The negative sign of the velocity at time t = 1s shows that the particle is moving in the negative x-direction.

- The speed ( s ( t )is the absolute value of velocity at time t = 1s:

                            s ( t ) = abs ( v ( t ) )

                            s ( 1 ) = abs ( v ( 1 ) )  

                            s ( 1 ) = abs ( -6 )

                            s ( 1 ) = 6 m/s

- The speed of the particle at time t = 0,

                            s ( t ) = abs ( -12 + 6t )

                            s ( 0 ) = abs (-12 + 6 (0) )  

                            s ( 0 ) = abs ( -12 )

                            s ( 0 ) = 12 m/s

- The speed of the particle at time t = 2,

                            s ( t ) = abs ( -12 + 6t )

                            s ( 2 ) = abs (-12 + 6 (2) )  

                            s ( 2 ) = abs (  0 )

                            s ( 2 ) = 0 m/s

- Hence, the speed of the particle decreases from s ( 0 ) = 12 m/s to s ( 2 ) = 0 m/s in the time interval t = 0 to t = 2 s.

- As the speed decreases as time increases over the interval t = 0 , t = 2 s the velocity v(t) also approaches 0, at time t = 2 s. s ( 2 ) = 0 m/s.

- We will develop an inequality when v (t) is positive:

                            v (t) = -12 + 6t > 0

                            6t > 12

                            t > 2

- So for all values of t > 2 the velocity of the particle is always positive.