IV. The table below shows how the speed of a car changes with time. 1 What is the acceleration of the car? please help! :(​

Answer:

prefix that means 1/100 = Centi

prefix that means 1,000 = Kilo

prefix that means 1/1000 = Milli

Explanation:

Question: A ship anchored at sea is rocked by waves that have crests 100 m apart the waves travel at 70m/S, at what frequency do the waves reach the ship?

Answer:

0.7 Hz

Explanation:

Applying,

v = λf............... Equation 1

Where v = velocity of the wave, f = frequency fo the wave, λ = wavelength of the wave

make f the subject of the equation

f = v/λ................. Equation 2

From the question,

Given: v = 70 m/s, λ = 100 m ( distance between successive crest)

Substitute these values into equation 2

f = 70/100

f = 0.7 Hz

Hence the frequency at which the wave reach the ship is 0.7 Hz

Answer:

Buoyant force exerted water​ =  0.196 Newton

Explanation:

Given:

Mass of sphere ball = 5 kg

Volume = 2 x 10⁻⁵

Find:

Buoyant force exerted water​

Computation:

Buoyant force exerted water​ = Gravity due to acceleration x volume of object x density of given liquid

Buoyant force exerted water​ = 9.8 x 2 x 10⁻⁵ x 1000

Buoyant force exerted water​ =  0.196 Newton

Explanation:

area=length(m) ×breadth(m) . The unit of area is expressed in terms of fundamental units m^2.thus it is derived unit

Answer:

C.) Longitudinal

Answer:

The primary nutrients are nitrogen, phosphorus and potassium.

The intermediate nutrients are sulfur, magnesium, and calcium.

The remaining essential elements are the micronutrients and are required in very small quantities.

Answer:

The primary nutrients are nitrogen, phosphorus and potassium.

The intermediate nutrients are sulfur, magnesium, and calcium.

The remaining essential elements are the micronutrients and are required in very small quantities.

Explanation:

Answer:

The answer is Speed=2m/s

Explanation:

S=D/T

S=10m/5s

S=2m/s

Answer:

Período del tambor: [tex]T = 0.25\,s[/tex], fuerza sobre la prenda: [tex]F \approx 80.852\,N[/tex], velocidad lineal del tambor: [tex]v \approx 10.053\,\frac{m}{s}[/tex], velocidad angular del tambor: [tex]\omega \approx 25.133\,\frac{rad}{s}[/tex].

Explanation:

La expresión tiene un error por omisión, su forma correcta queda descrita a continuación:

"Una prenda de 320 gramos de ropa gira en el interior de una lavadora si dicha lavadora tiene un radio de 40 centímetros y gira con una frecuencia de 4 hertz. Halle a) el período, b) la velocidad angular, c) la fuerza con la que gira la prenda y d) la velocidad lineal de la lavadora."

El tambor gira a velocidad angular constante ([tex]\omega[/tex]), en radianes por segundo, lo cual significa que la prenda experimenta una aceleración centrífuga ([tex]a[/tex]), en metros por segundo al cuadrado. En primer lugar, calculamos el período de rotación del tambor ([tex]T[/tex]), en segundos:

[tex]T = \frac{1}{f}[/tex] (1)

Donde [tex]f[/tex] es la frecuencia, en hertz.

([tex]f = 4\,hz[/tex])

[tex]T = \frac{1}{4\,hz}[/tex]

[tex]T = 0.25\,s[/tex]

Ahora determinamos la fuerza aplicada sobre la prenda ([tex]F[/tex]), en newtons:

[tex]F = m\cdot a[/tex] (2)

[tex]F = \frac{4\pi^{2}\cdot m \cdot r}{T^{2}}[/tex] (2b)

Donde:

[tex]m[/tex] - Masa de la prenda, en kilogramos.

[tex]r[/tex] - Radio interior del tambor, en metros.

([tex]m = 0.32\,kg[/tex], [tex]r = 0.4\,m[/tex], [tex]T = 0.25\,s[/tex])

[tex]F = \frac{4\pi^{2}\cdot (0.32\,kg)\cdot (0.4\,m)}{(0.25\,s)^{2}}[/tex]

[tex]F \approx 80.852\,N[/tex]

La velocidad lineal de la lavadora es:

[tex]v = \frac{2\pi\cdot r}{T}[/tex] (3)

([tex]r = 0.4\,m[/tex], [tex]T = 0.25\,s[/tex])

[tex]v = \frac{2\pi\cdot (0.4\,m)}{0.25\,s}[/tex]

[tex]v \approx 10.053\,\frac{m}{s}[/tex]

Y la velocidad angular del tambor de la lavadora:

[tex]\omega = \frac{2\pi}{T}[/tex]

([tex]T = 0.25\,s[/tex])

[tex]\omega = \frac{2\pi}{0.25\,s}[/tex]

[tex]\omega \approx 25.133\,\frac{rad}{s}[/tex]

Answer:

A.

Explanation:Object b will get a negative charge .

Answer:

Kinetic Energy

Explanation:

Answer:

The mass of the dog food added is 9.03 kg

Explanation:

Given;

mass of the shopping cart, m₁ = 14.5 kg

let the mass of the bag added = m₂

the force applied, F = 12 N

initial velocity of the cart-bag system, u = 0

distance traveled by the system, d = 2.29 m

time of motion of the system, t = 3.0 s

The acceleration of the system is calculated as;

[tex]d = ut + \frac{1}{2} at^2\\\\2.29 = 0 + (\frac{1}{2} \times 3^2)a\\\\2.29 = 4.5 a\\\\a = \frac{2.29}{4.5} \\\\a = 0.51 \ m/s^2[/tex]

The total mass of the system (M) is calculated as follows;

F = Ma

M = F/a

M = (12)/(0.51)

M = 23.53 kg

The mass of the dog food added is calculated as;

m₂ = M - m₁

m₂ = 23.53 kg - 14.5 kg

m₂ = 9.03 kg

Solution :

It is given that the lamp glows brightly for a shorter period of time when the switch is closed on it the switch is put on. But after the some time the lamp goes off and it stops working.

This is because as soon as we on the switch, the current start flowing to the lamp which makes the filament of the lamp to glow, but due to some issue, the current stop flowing even when the switch is on and this stops the lamp from glowing and hence the lamp does not work.

Assume R is measured in meters (m) and M in kilograms (kg). Then

R ² / (GM) = [m]² / ([N•m²/kg²] [kg]) = m•kg / N = m•kg / (kg•m/s²) = s²

so t ² is indeed proportional to R ²/(GM).

Answer:

a.

[tex]F=G\cdot\dfrac{M \cdot m}{r^{2}}[/tex]

b.

[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}}[/tex]

c.

[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}} \approx 1.144 \times 10^{13} \ N[/tex]

d. The force of gravity acting on the satellite is approximately 1.144 × 10¹³ N

Explanation:

a. The formula for finding the force of gravity, F, acting object on an object is given as follows;

[tex]F=G\cdot\dfrac{M \cdot m}{r^{2}}[/tex]

Where;

F = The force acting between the Earth and the object

G = The gravitational constant = 6.67430 × 10⁻¹¹ N·m²/kg²

M = The mass of the Earth = 5.97 × 10²⁴ kg

m = The mass of the object

r = The distance between the center of the Earth and the object

b. Finding the gravitational force, 'F', between the Earth and the scientific satellite, we have;

The given mass of the satellite, m = 1,300 kg

The distance between the center of the Earth and the center of the satellite,  r = The length of the radius of the Earth + The height of orbit of the satellite

The given height of orbit of the satellite, h = 200 km

∴ r = R + h = 6,378 km + 200 km = 6,578 m

Therefore, by plugging in the values, we get;

[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}}[/tex]

c. Solving the above equation gives;

[tex]F=6.67430 \times 10^{-11} \dfrac{N \cdot m^2}{kg^2} \times \dfrac{5.97 \times 10^{24} \ kg \times 1,300 \ kg}{(6,578 \ m)^{2}} \approx 1.144 \times 10^{13} \ N[/tex]

d. The force of gravity acting on the satellite, F ≈ 1.144 × 10¹³ Newton

Answer:

Friction Opposes Motion of an Object.

Now

To get the Net force that Moves an Object and causes acceleration....You subtract the Frictional force

Net force = Pushing Force - Frictional Force

Recall

Net Force; F=Ma

Ma = P - Fr

Now the question asked for How Much force Must be applied to Maintain a Constant velocity.

In a Constant Velocity Motion... Acceleration do not change... Its Zero

So Putting this into the formula above

M(0) = P - Fr

0=P - Fr

Fr = P.

This means

That The force needed to keep this object Moving at Constant Velocity Must be equal to its Frictional Force

Since Frictional Force; Fr =223N

The Applied Force(Pushing Force) Must be equal to 223N too.

Answer:

the angular acceleration of the car is 1.5 rad/s²

Explanation:

Given;

initial angular velocity, [tex]\omega_i[/tex] = 10 rad/s

final angular velocity, [tex]\omega_f[/tex] = 25 rad/s

time of motion, t = 10 s

The angular acceleration of the car is calculated as follows;

[tex]a_r = \frac{\omega_f - \omega_i }{t} \\\\a_r = \frac{25-10}{10} = 1.5 \ rad/s^2[/tex]

Therefore, the angular acceleration of the car is 1.5 rad/s²

Answer:

only long-range force that affects all particles is the gravitational force.

Explanation:

In nature there are four fundamental forces: nuclear, weak, gravitational and electrical.

The last two are long-range, that is, the forces are zero for infinite distances, the current gravitational on all the particles and the electric one acts on the charged particles, without the chosen charge it is zero, the forces is also zero.

Consequently the only long-range force that affects all particles is the gravitational force.

Answer:

Explanation

The most famous and common example is Christmas tree lights. You can't tell easily by looking at them whether they are in series or parallel. But you sure know the difference when one of them burns out. When that happens, the whole string goes dead. No matter what you do (other than find out which bulb burned out) will not fix the problem.

Another example is anything that is temperature controlled. For example a furnace is controlled by a thermostat. When the room temperature reaches a certain point, the thermostat is constructed in a certain way so that it forms an open circuit and no current can flow through it. The furnace motor turns off and the furnace stops pumping hot air into a room.

Answer:

4.71m/s

Explanation:

Average speed = Total distance travelled ÷ Total time taken.

80/17=4.71

4.71m/s

Answer:

In this question,

                                                = 80m.

                                     = 17s

So, Average speed would be 80 ÷ 17

= [tex]\frac{80}{17}[/tex]

 = 4.71 m/s. or 4.71 meter per second.

Answer:

A...................................

The force of the wall on the car and the car on the wall are equal is true about Newton’s Third Law. Option A is the correct answer.

According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction. This means that if the car hits the wall, there will be a force exerted by the car on the wall, and an equal and opposite force exerted by the wall on the car. Option A is the correct answer.

The forces involved in the interaction between the car and the wall are equal in magnitude but opposite in direction, as dictated by Newton's Third Law. Newton's Third Law of Motion states that for every action, there is an equal and opposite reaction. This means that when an object exerts a force on another object, the second object exerts a force of equal magnitude but in the opposite direction on the first object.

Learn more about Newton here:

https://brainly.com/question/28171613

#SPJ2

The complete question is, "In the picture below, a car hits a wall. Using what you know about Newton’s Third Law, which is true?

a. The force of the wall on the car and the car on the wall are equal

b. The force of the wall on the car is greatest

c. The force of the car on the wall is greatest

d. There is not enough information to tell"

Explanation:

i. Vi=24

Vf=0

t= -2

a=vf-vi/t =0-24/12 = -2m/S2

ii. F=ma = 1500×-2= -3000 N

Answer:

by using v = u + at equation we can find "a"

14 = 60 - 2.7a

2.7a = 60 - 14

2.7a = 46

decceleration = 17.03

Answer:

Also, as stream depth increases, the hydraulic radius increases thereby making the stream more free flowing. Both of these factors lead to an increase in stream velocity. The increased velocity and the increased cross-sectional area mean that discharge increases.

Answer:

there it is fella tried on ma own observation

Answer:

Cause you are their family and they need you. Do you wanna be someone who abandoned them or be the one they look to and dont dislike into adulthood. Also you signed onto this job no one said it was going to be easy but you making your life easier but harder for someone who is just a kid who still needs you.

Explanation:

1. if an object sjnks in one liquid and floats on another liquid it implies that the density of second liquid is greater than the density of first liquid