A box is at rest on a table. What can you say about the forces acting on the box?
The upward
and the downward
are balanced.

The acceleration of the snowboarder is 0.466 m/s².

Acceleration is the rate of change of velocity.

To calculate the acceleration of the snowboarder, we use the formula below.

Formula:

Where:

From the question,

Given:

Susbtitute these values into equation 1

Hence, the acceleration is 0.466 m/s².

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Answer:

ahhahsvdjdjkd

Explanation:

sbnwtauajababcahajanavgzuz

Answer:

the f is 1.3 m

Explanation:

Given that

An object 2.0m is from the concave mirror

And, the image is 4.0m from the mirror

We need to find out the f

So,

u = -2 m

v = -4 m

1 ÷ f = 1 ÷ v + 1 ÷ u

= (-1 ÷ 4) + (-1 ÷ 2)

= -3 ÷4

f = -4 ÷ 3

= |1.3| m

Hence, the f is 1.3 m

Answer: [tex]0.1066\ psi, 15.611\ kN[/tex]

Explanation:

Given

the height of the tank is [tex]h=10\ cm[/tex]

The diameter of the tank is [tex]d=520\ cm[/tex]

Density of solution [tex]\rho=0.75\ g/cm^3\ or\ 750\ kg/m^3[/tex]

[tex](a)[/tex] Water pressure at the bottom of the tank is

[tex]P=\rho gh\\P=750\times 9.8\times 0.1\\P=735\ Pa[/tex]

[tex]1\ Pa=0.000145038\ psi[/tex]

[tex]\Rightarrow P=735\ Pa\ or\ 0.1066\ psi[/tex]

[tex](b) \text{Average force on the bottom is the product of pressure and area of the base}[/tex]

[tex]F_{avg}=735\times \pi \cdot (\frac{520}{200})^2\\\\F_{avg}=735\times 3.142\times 6.76\\F_{avg}=15,611.34\ N\ or\ 15.611\ kN[/tex]

Answer:

Therefore The hypothesis is not plausible Because

[tex]P.E \geq P.E_e[/tex]

[tex]P.E=12.435*10^{-5}J \geq 1.4*10^{-7}J[/tex]

Explanation:

From the question we are told that:

Mass [tex]M=0.45mg[/tex]

Jump height [tex]h=3.5cm \approx 0.03m[/tex]

Percentage strain [tex]\mu =100\%[/tex]

Generally the equation for potential energy is mathematically given by

[tex]P.E=mgh[/tex]

[tex]P.E=0.45*10^{-3}*9.8*0.03[/tex]

[tex]P.E=12.435*10^{-5}J[/tex]

Generally the equation for Elastic potential energy is mathematically given by

[tex]P.E_e=0.5*stress*strain *volume[/tex]

[tex]P.E_e=0.5*1*2*10^6*1*1.4*10^{-4*(10^-3)^3}[/tex]

[tex]P.E_e=1.4*10^{-7}J[/tex]

The hypothesis is only plausible at conditions where

[tex]P.E\leq P.E_e[/tex]

Therefore The hypothesis is not plausible Because

[tex]P.E \geq P.E_e[/tex]

[tex]P.E=12.435*10^{-5}J \geq 1.4*10^{-7}J[/tex]

Answer: A) as speed changes, wavelength changes and frequency remains constant

Explanation: edg test

Answer:

Node:-

A node is a point along a standing wave where the wave has minimum amplitude. For instance, in a vibrating guitar string, the ends of the string are nodes. By changing the position of the end node through frets, the guitarist changes the effective length of the vibrating string and thereby the note played.

Anti Node:-

An antinode is simply a point along a medium which undergoes maximum displacement above and below the rest position.

Wavelength:-

The total length of circle.

or

Wavelength can be defined as the distance between two successive crests or troughs of a wave.

Answer:

ii. N

Explanation:

N is not laterally symmetrical. so it shows inversion in a mirror.

Answer:

ii) N

hope it is helpful to you

Answer:

B.KEEN EYESIGHT thats for predators !

Explanation:

Herbivores have multiple stomachs to process tough foods, coats to blend in so they don't get eaten, and tough teeth for chewing on the plants. THEY DO NOT HAVE PREY, THEY ARE THE PREY.

Answer:

Number 19

10 x 7.2

72N

Explanation:

20.

1.66 x 7.2

11.9

Which is approximately

12N

Answer:

the  correct  answers B

It needs to be moving horizontally at a high speed, in order for the satellite to stay in orbit once it's released from the rocket.

Explanation:

In this exercise, let's look for the solution to the problem before reviewing the statements to find out which one is correct.

When the satellite is in orbit it must comply with Newton's law

          F = m a

the force of the universal attractive force and the acceleration is centripetal

          a = v² / r

we substitute

         - G m M /r² = -m v² / r

            v² = G M / r

            v = [tex]\sqrt{\frac{GM}{r} }[/tex]

When analyzing this expression, the satellite is in a stable orbit because its centripetal acceleration creates a change in the direction of the velocity in such a way that it remains in an orbit

when reviewing the  correct  answers B

The total work done by this force is 28.9-1457.86 N m.

Total work is the sum of all the energy expended in completing a job or task. It is the amount of effort and energy expended to accomplish a goal or complete a task. Total work can be calculated by adding up all of the individual components of the job or task, such as time, effort, and materials.

The total work done by this force can be calculated using the formula W = F * Δx, where F is the force vector, and Δx is the displacement vector. In this case, the total work done is:
W = (4.70-379, 2.09) * (4.25, 3.69-245) = (19.9-944.81, 8.00-513.05) N m
Therefore, the total work done is 28.9-1457.86 N m.

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Answer:

24.4

Explanation:

you would add all the numbers , 25.5+24.4+24.2= 73.1/3=24.36 and you round up you get 24.4

===================================================

Explanation:

vi = 5 and vf = 8 are the initial and final velocities respectively. The change in time is t = 20 seconds.

So,

x = 0.5*(vi + vf)*t

x = 0.5*(5+8)*20

x = 130 meters

represents the distance traveled. The first equation shown above is one of the four kinematics equations.

Based on the data provided about the collision;

The magnitude of the impulse due to the collision of the vehicle with the wall is calculated using the formula of the impulse on an object given below as follows:

Impulse = change in momentum

Change in momentum = m(v - u)

where;

m is mass

v is the final velocity

u is the initial velocity

Impulse = 1759 (12 - 22)

Impulse = 21105.8 kg.m/s

The average force on the automobile will be:

Force = impulse/time

Force = 21105.8 / 0.11

Force = 191870.91 N

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Answer:

i think its Hydrodynamics

Explanation:

because hydro is fluid

If deceleration = 5 m/s²

then acceleration = -5 m/s²

According to third law of motion,

= v² = 2as + u²

= v² = 2×-5×10 + 27²

= v² = -100 + 729

= v =√ 629

= 25.08 m/s

This might help as well:

https://www.mathsisfun.com/physics/light.html

Answer: A. The Sun

Explanation: A. The Sun makes water move through the water cycle

Answer:

c. more redder

def not blue cuz thats hottest and def not luminous

Answer:

it is c... redder

Explanation:

because i just look it up on googl3

Answer:

1.08 x 10^5 J

Explanation:

First, apply the formula V = IR , where V is the voltage, I is the current and R is the resistance.

R -> 20 ohm (Given)

I -> 3A (Given)

Hence, V = 3 x 20 = 60V (60 volts)

Next, apply the formula E = Pt , where E is the energy produced, P is the power and t is the time in seconds

Furthermore, since P = IV (power = current x voltage)

The formula can be re-expressed as E = VIt

t -> 10 minutes = 600 seconds (don't forget to convert time to seconds!)

Therefore, energy = (60 x 3 x 600) J

                              = 108000 J

                               = 1.08 x 10^5 J (in scientific notation)

Hea produced in the resistance = energy produced in the resistor

                                                      = 1.08 x 10^5 J

Answer:

666.7 W

Explanation:

Power = P = W/t = 4000 J/6 s = 666.7 W (watts)

Answer: it always opposes the directions of motion is the right answer

Explanation:

Answer:

The difference between mass and weight is that mass is the amount of matter in a material, while weight is a measure of how the force of gravity acts upon that mass.

Explanation:

Answer:

A. 9.203 × 10⁻¹⁰ kg

B.  2.82 × 10¹⁰ V/m

C. 5.56 × 10³ V/m

Explanation:

A. Calculate the mass (in kg) of the hovering droplet if its radius is measured to be 65 um.

We know density, ρ = mass of oil drop, m/volume of oil drop, v

m = ρv

ρ = 800 kg/m³ and v = 4πr³/3 (ince the oil drop is a sphere) where r = radius of oil drop = 65 μm = 65 × 10⁻⁶ m.

So, m = ρv = ρ4πr³/3

= 800 kg/m³ × 4π(65 × 10⁻⁶ m)³/3

= 800 kg/m³ × 4π274625 × 10⁻¹⁸ m³/3

= 878800000π × 10⁻¹⁸ kg/3

= 2760831623.97 × 10⁻¹⁸ kg/3

= 920277207.99 × 10⁻¹⁸ kg

= 9.203 × 10⁻¹⁰ kg

B. Calculate the electric field (in V/m) required to counter the weight of the above droplet if it carries a charge q = 2e.

Since the electric force F = qE where q = charge on oil drop  = 2e where e = electron charge = 1.602 × 10⁻¹⁹ C and E = electric field equals the weight of the oil drop W = mg where m = mass of oil drop = 9.203 × 10⁻¹⁰ kg and g = acceleration due to gravity = 9.8 m/s².

So, F = W

qE = mg

E = mg/q

E = mg/2e

substituting the values of the variables into the equation, we have

E = 9.203 × 10⁻¹⁰ kg × 9.8 m/s²/(2 × 1.602 × 10⁻¹⁹ C)

E = 90.1894 × 10⁻¹⁰ kg-m/s²/3.204 × 10⁻¹⁹ C

E = 28.15 × 10⁹ V/m

E = 2.815 × 10¹⁰ V/m

E ≅ 2.82 × 10¹⁰ V/m

C. In the second trial, the hovering droplet has a mass of 10-15 kg. It carries a charge of q = 11e. calculate the electric field (in V/m) required to counter the weight of this droplet.

Since F = W

qE = mg

E = mg/q

E = mg/11e

E = 10⁻¹⁵ kg × 9.8 m/s²/(11 × 1.602 × 10⁻¹⁹ C)

E = 9.8 × 10⁻¹⁵ kg-m/s²/17.622 × 10⁻¹⁹ C

E = 0.556 × 10⁴ V/m

E = 5.56 × 10³ V/m

Answer:

-0.0069 rad/s

Explanation:

The change in angular velocity of the rod can be determined by using the conservation of angular momentum equation: Lf = Li.

The initial angular momentum of the rod is Li = Iwo and the final angular momentum of the rod is Lf = Iwf.

Plugging in the given values, we get:

(1.05 x 10^-3 kg•m^2)(0.606 rad/s) + (5 x 10^-3 kg)(0.609 m)^2(wf) = (1.05 x 10^-3 kg•m^2)(wo).

Solving for wf, we get wf = -0.0069 rad/s.

Therefore, the change in angular velocity of the rod is -0.0069 rad/s.

Answer:

a. E = 2452.5 J = 2.45 KJ

b. P.E = 981 J = 0.98 KJ

c. K.E = 1471.5 J = 1.47 KJ

d. vf = 14 m/s

Explanation:

a.

At the highest point the total energy is equal to the potential energy of the object:

E = Potential Energy = mgh

where,

E = Total energy =?

m = mass of object = 25 kg

g = acceleration due to gravity = 9.81 m/s²

h = height = 10 m

Theerfore,

E = (25 kg)(9.81 m/s²)(10 m)

E = 2452.5 J = 2.45 KJ

b.

P.E = mgh

where,

h = 4 m

Therefore,

P.E = (25 kg)(9.81 m/s²)(4 m)

P.E = 981 J = 0.98 KJ

c.

First, we will find the velocity at 4 m by using the third equation of motion:

[tex]2gh = v_f^2-v_i^2[/tex]

where,

h = height lost = 10 m - 4 m = 6 m

vf = final speed = ?

vi = initial speed = 0 m/s

Therefore,

[tex]2(9.81\ m/s^2)(6\ m) = v_f^2-(0\ m/s)^2\\v_f = \sqrt{117.72\ m^2/s^2} \\v_f = 10.85\ m/s[/tex]

Hence, the kinetic energy will be:

[tex]K.E = \frac{1}{2} mv_f^2\\\\K.E = \frac{1}{2} (25\ kg)(10.85\ m/s)^2[/tex]

K.E = 1471.5 J = 1.47 KJ

d.

We will find the velocity at the bottom by using the third equation of motion:

[tex]2gh = v_f^2-v_i^2[/tex]

where,

h = height lost = 10 m

vf = final speed = ?

vi = initial speed = 0 m/s

Therefore,

[tex]2(9.81\ m/s^2)(10\ m) = v_f^2-(0\ m/s)^2\\v_f = \sqrt{196.2\ m^2/s^2} \\[/tex]

vf = 14 m/s

To solve this we must be knowing each and every concept related to interval. Therefore, values of the displacement of the string at x = 5.0 m at 1 s intervals from t = 0 s to t = 6 s can be listed as below.

An interval is any set of numbers that fall between two distinct integers. This range includes all real values that fall between those two. All real numbers that lie in any 2 of a set's numbers are collectively referred to as an interval within mathematics.

For instance, the set of integers x fulfilling 0 x 1 is the interval consisting of 0, 1, and all numbers in between. a snapshot graph of wave on a string moving toward one another at 1 m/s at time t = 0 s.

At time t = 0,Y₀= 0 cm

At At time t =1, Y₁= 0 cm

At time t = 2, Y₂= 0 cm

At time t =3, Y₃= 0 cm

At time t =4, Y₄= 0 cm

At time t = 5 ,Y₅=0 cm

At time t = 6, Y₆= 0 cm

Therefore, values of the displacement of the string at x = 5.0 m at 1 s intervals from t = 0 s to t = 6 s can be listed as above.

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Answer:

The answer is "530 nm".

Explanation:

[tex]d=0.180 \ nm=0.180 \times 10^{-3} \ m\\\\L= 5.05\ m\\\\\Delta y= 1.62\ cm= 1.62 \times 10^{-2} \ m\\\\[/tex]

Using formula:

[tex]\Delta y=\frac{\lambda L}{d}\\\\\lambda =\frac{d\Delta y}{L}\\\\[/tex]

  [tex]=\frac{0.180 \times 10^{-3} \times 1.62 \times 10^{-2} }{5.05}\\\\=5.30 \times 10^{-9} \times \frac{1 \ nm}{10^{-9} \ m}\\\\=530 \ nm[/tex]