how to find initial velocity without acceleration

Answer:

the slope of velocity-time graph represent an object acceleration

Answer:

Homogenous Mixture

Explanation:

The composition in paint is uniform so every part of it looks the same throughout the paint.

Answer:

d and c are the answers :)

Answer:

You should try your best to answer the question.

D)

Explanation:

Coz the position changes with time but the position change is not constant throughout time So, D) does not have constant velocity

Sorry your tired, ill try to make it worth reading.

One day I walked outside, I looked around. THEN BAM a wasp lands right on my face. Its hideously sharp stinger impaled my flesh. I could almost hear it going deeper and deeper. I screamed and ran around in circles until it finely flew away. As I started walking to the door I could feel my heart pules where the thing had stung. I opened the door and went into the bathroom. Right in the middle of my face just above my nose was a red dot, with a slightly less red ring around it, like an ateroid impact. I sat down and went over the entire event five times. I decided to go back outside, as I walked around the corner, another wasp landed on me.

Force = mass x acceleration:

Force = 13.2 kg x 9.4 m/s

Force = 124.08

Force = 124 Newtons

The force will be  = 124 N              

Newton's second law is a quantitative description of the changes that a force can produce on the motion of a body. It states that the time rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it.

according to newton's second law of motion

Force = mass * acceleration

given

mass = 13.2 kg

acceleration = 9.4 m/s

force = 13.2 * 9.4

        = 124.08 N

Rounding off answer to  the appropriate number of significant figures

force = 124 N

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The process of charging an object without touching it is called induction charging.

The process of charging a sphere by touching it with a charged rod is called charging by conduction..

When an uncharged object brings into the contact of the charged object but not touching the object directly, the uncharged object gets charged. This process of charging is called Induction Charging.

When an uncharged object brings in to the contact of the charged object by touching it directly, the uncharged object gets charged, This process of charging is called Conduction Charging.

When a charged rod touches the metal sphere, sphere gets same charged as the rod. The process of charging a sphere by touching it with a charged rod. is called conduction charging.

For more details about the conduction and induction charging, follow the link given below.

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Answer:

induction and conduction or 3,2 or c and b

Explanation:

edge 2022

Answer:

Plan B.

Because flexibility is best improved by stretching.

Explanation:

Improving and increasing flexibility is done by having stretching sessions daily which maintains and widens the range of motion in the joints and stretches muscles.

Answer:

The value of [tex]\theta=63.43^0[/tex]

Explanation:

Range of projectile [tex]\mathrm{R}=\frac{\mathrm{u}^{2} \cdot \sin 2 \theta}{\mathrm{g}}[/tex]

Height of projectile [tex]\mathrm{h}=\frac{\mathrm{u}^{2} \cdot \sin ^{2} \theta}{2 \cdot \mathrm{g}}[/tex]

Here we have

[tex]R=2h\\\\\frac{\mathrm{u}^{2} \cdot \sin 2 \theta}{\mathrm{g}}=2\frac{\mathrm{u}^{2} \cdot \sin ^{2} \theta}{2 \cdot \mathrm{g}}\\\\\sin 2 \theta=\sin ^{2} \theta\\\\2\sin \theta\cos \theta=\sin ^{2} \theta\\\\\tan \theta=2\\\\\theta=63.43^0[/tex]

Range be R and height be h

[tex]\boxed{\sf R=\dfrac{u^2sin2\theta}{g}}[/tex]

[tex]\boxed{\sf h=\dfrac{u^2sin^2\theta}{2g}}[/tex]

ATQ

[tex]\\ \sf\longmapsto R=2h[/tex]

[tex]\\ \sf\longmapsto \dfrac{u^2sin2\theta}{g}=\dfrac{2u^2sin^2\theta}{2g}[/tex]

[tex]\\ \sf\longmapsto sin^2\theta=sin2\theta[/tex]

[tex]\\ \sf\longmapsto sin^2\theta=2sin\theta cos\theta [/tex]

[tex]\\ \sf\longmapsto \dfrac{sin^2\theta}{sin\theta cos\theta=2[/tex]

[tex]\\ \sf\longmapsto \dfrac{sin\theta}{cos\theta}=2[/tex]

[tex]\\ \sf\longmapsto tan\theta=2[/tex]

[tex]\\ \sf\longmapsto \theta=tan^{-1}(2)[/tex]

[tex]\\ \sf\longmapsto \theta=63.4°[/tex]

I am familiar with Russian Mountaineering

Answer:

2,500g is the answer yes

Answer:

1 N

Explanation:

1) You'll just have to sum the force vectors, and since they are opposite, you'll calculate the difference of their magnitudes, in absolute value.

Ftot = |F1 - F2| = 6N - 5N = 1N

2)The direction will be the same as the greater force's direction.

Current Forecast is:

Sunny with a temp of 67 deg F, as of rn there is no clouds in the sky

Location, Oklahoma USA in the NE corner

Developing Psychological Theories: Mastery Test

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Hope this helps :)

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So, your science teacher has given your class the classic "mousetrap car" assignment: to make, design, and build a small vehicle powered by the snapping action of a mousetrap to make your car travel as far as possible. If you want to come out ahead of all the other students in your class, you'll need to make your car as efficient as possible so you can squeeze every last inch out of your "car". With the right approach, it's possible to streamline your car's design for maximum distance using only common home materials. You could also buy a mousetrap car kit from any craft store and skip wondering if it will work.

Use large rear wheels. Large wheels have greater rotational inertia than small wheels. In practice, this means that once they start rolling, they're harder to stop rolling. This makes large wheels perfect for distance-based contests — theoretically, they'll accelerate less quickly than smaller wheels, but they'll roll much longer and they'll travel a greater distance overall. So, for maximum distance, make the wheels on the drive axle (the one the mousetrap is tied to, which is usually the rear one) very large. The front wheel is a little less important — it can be large or small. For a classic drag racer look, you'll want big wheels in the back and smaller ones in front.

Use thin, light wheels. Thinner wheels have less friction and may go farther if the distance is what you want or need with your mousetrap racer. It's also important to take the weight of the wheels themselves into account — any unneeded weight will ultimately slow your car down or lead to added friction. In addition, it's worth noting that wide wheels can even have a small negative effect on the car's drag due to air resistance. For these reasons, you'll want to use the thinnest, lightest wheels available for your car.

Old CDs or DVDs work fairly well for this purpose — they're large, thin, and extremely light. In this case, a plumbing washer may be used to reduce the hole size in the middle of the CD (to fit the axle better).

If you have access to old vinyl, these also work extremely well, though they may be too heavy for the smallest mousetraps.

Use a narrow rear axle. Assuming your car is a rear-wheel-drive car, each time your rear axle turns, the rear wheels turn. If your rear axle is extremely skinny, your mousetrap car will be able to turn it more times for the same length of string than it would if it were wider. This translates to turning your rear wheels more times, meaning greater distance! For this reason, it's a wise idea to make your axle out of the skinniest material available that can still support the weight of the frame and wheels.

Narrow wooden dowel rods are a great, easily-accessible choice here. If you have access to thin metal rods, these are even better — when lubricated, they usually have less friction.

Create traction by giving the edges of the friction of the wheels. If the wheels slip against the ground when the trap is sprung, energy is wasted — the mousetrap works to make the wheels turn, but you don't get any extra distance. If this happens with your car, adding a friction-inducing material to the rear wheels may reduce their slippage. To keep your weight requirements down, use only as much as is necessary to give the tips of the wheels some grip and no extra. Some suitable materials are:[1]

Electrical tape

Rubber bands

Additionally, placing a piece of sandpaper under the rear wheels at the start line can reduce slippage as the car begins to move (when it is most likely)

Answer:

V = 4/3 R^3

V2 / V1 = (R2 / R1)^3 = 2^3 = 8

8.00

The volume of water in the larger balloon is 8 times greater than in the smaller balloon

We'll begin by calculating the volume of each balloon.

V = 4/3 πr³

V = 4/3 × 3.14 × 3³

V = 4/3 πr³

V = 4/3 × 3.14 × 6³

Finally, we shall determine how much greater the larger balloon is to the smaller balloon

Greatness => large / small

Large / small = 904.32 / 113.04

Large / small = 8

Cross multiply

Therefore, the larger balloon is 8 times greater than the smaller balloon.

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Answer:

So correct sentence is:

So correct sentence is:

So correct sentence is:

So correct sentence is:

Answer:

Reaction force = Weight

the force exerted by A on B is equal.

M1a = M2a  

Explanation: I think the answer is right if wrong message me.

Explanation:

hope this may help you

have a nice day

Answer:

Inside the film the wavelength will be λ/n

For constructive interference to occur the film must be λf/4 thick where λf is the wavelength of the light in the film - there will be a 180 degree phase shift at the water/film interface since the index of refraction of the film is greater than that of water - and the light has to travel λ/2 inside the film for constructive interference to occur

280 nm / 1.6 * 4 = 700 nm is the greatest wavelength allowed

Note that 700 nm is also the upper wavelength of the visible spectrum

Answer:

because of the gravity of the earth

Answer:

We know that ΔK = Kf - Ki = 1/2 m Vf^2 - 1/2 m Vi^2 = 1/2m(Vf^2-Vi^2) = 1/2 m ΔV^2.

The mass remains the same, just calculate the difference of squared velocities and multiply it by half of the mass.

Answer:

density can be written as:

d = F / V  (weight density) or M / V (mass density)

Since F in the first equation can be written as

B = d V where B is the buoyant force and is proportional to the

weight of the liquid displaced

For a given volume the weight of the liquid displaced is proportional to the density and hence the buoyant force

[tex]\\ \sf\Rrightarrow F_{net}=F_1+F_2+F_3+F_4[/tex]

[tex]\\ \sf\Rrightarrow F_{net}=20N+20N+10N-10N[/tex]

[tex]\\ \sf\Rrightarrow F_{net}=40N[/tex]

Answer:

I think you ask for another way of writing the joule (J). It can be expressed as the product between a force applied to a body and the deployment of the body (not necessarily caused by that force applied), so N*m (newton * meter).

The quantity work—which has units of energy—was defined, another unit for energy, equivalent to the joule (j), is Newton ·meter (N· m).

The quantity with a constant magnitude that is used to measure the magnitudes of other quantities of the same type is referred to as a unit in physics.

Unit of work is that of energy, that is, Joule (J).

Again, work = force × displacement

Unit of work = Unit of force × Unit of displacement = Newton . meter (N.m).

Hence, the quantity work—which has units of energy—was defined, another unit for energy, equivalent to the joule (j), is Newton ·meter (N· m).

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The strength of the magnetic field is increased by adding more coils to the wire and placing a metal bar within the coils.

The given problem is based on the concept and fundamentals of the magnetic strength in a current-carrying coil of wire. The current-carrying wire in a coiled form is known as Solenoid.

And the mathematical expression for the magnetic field in a solenoid is,[[tex]B = \dfrac{\mu_{0} \times N \times i}{L}[/tex]

Here,

B is the strength of the magnetic field.

[tex]\mu_{0}[/tex] is the magnetic permeability of free space.

N is the number of coils.

The strength of the magnetic field will increase by increasing the number of coils and placing the metallic slab (it will increase magnetic permeability) within the coils.

Thus, we can conclude that the strength of the magnetic field is increased by adding more coils to the wire and placing a metal bar within the coils.

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Answer:

C

Explanation:

The distance traveled by object 1 is greater than the distance traveled by object 2.

The shortest distance between two points is a straight line. More distance is covered when the path of motion is curved.

The distance traveled by each object at different time is compared as follows;

[tex]at \ t_ A , \ \ \ d_1>d_2 \ \ \ (d_2 = 0)\\\\at \ t_B, \ \ d_1 > d_2 \\\\at \ t_C, \ \ \ d_2 > d_1 \\\\at \ t_D \ \ \ d_1 = d_2\\\\at \ t_f \ \ \ d_2 > d_1 \ \ \ (d_1 = 0)[/tex]

The major distinguishing distance traveled by each occurs at [tex]t_B[/tex] and [tex]t_C[/tex].

[tex]d_{(t_B)} > d_{(t_C)} \\\\d_1 > d_2[/tex]

Thus, we can conclude that the distance traveled by object 1 is greater than the distance traveled by object 2.

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