Which friction static friction, kinetic friction, or air friction? Can be more than one response.
Gripping friction.
kinetic friction
air friction
static friction

Answer:

B.) decreased attention span

Explanation:

took the test and got 100%

Answer

An object's temperature is the best approximation of the kinetic energy of the particles. When we measure an object's temperature, we measure the average kinetic energy of the particles in the object. The higher the temperature, the faster the molecules of the substance move, on the average

The measure of the warmth or coldness of a substance or an object with reference to some standard value is termed as temperature.

It is a measure of the coldness or hotness expressed in terms of any of the many arbitrary scales and suggesting the direction in which the heat energy will flow spontaneously, that is, from a hotter body to a colder body.

Similar to pressure or density, temperature is an intensive property, which is independent of the quantity of matter being considered.

Thus, temperature refers to a physical quantity, which expresses the extent of coldness or hotness of a substance.

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Answer:

I think it's matter

Explanation:

Because everything is made of matter

Answer:

Is there any thing miss from the question

Answer:

1) t_total = 140 min ,  v = 126.7 m / min ,  2)  v = 0.628 m / s ,

3) v = 0.592 m / s ,  θ = -39.3 º

Explanation:

Here we have some short kinematics problems

1) d₁ = 4200 m at T1 = 25 min

    d₂ = 4.54 km at t2 = 1 h

    d₃ = 9000 m at t3 = 55 min

of the statement the direction of each route is the same

Let's reduce the distances to the SI system

     d₂ = 4.54 km (1000 m / 1 km) = 4540 m

     t₂ = 1 h (60 min / 1 h) = 60 min

Total time is

    t_total = t1 + t2 + t3

   t_total = 25 + 60 +55

   t_total = 140 min

the distance is

    d_total = 4200 + 4540 +9000

     d_total = 17740 m

with movement it is in one dimension and all displacement is in the same direction, distance equals displacement.

Speed ​​is

       v = 17740/140

       v = 126.7 m / min

in the direction of travel

Speed ​​is the modulus of speed

        | v | = 126.7 m / min

2) the angular velocity is

        w = θ / t

In this exercise

        t = 2 min (60 s / 1min) = 120 s

        θ = 30 rev (2π rad / 1 rev) = 60π rad

         w = 60π / 120

         w = 1.57 rad / s

linear velocity  is

        v = w r

        v = 1.57 0.40

        v = 0.628 m / s

velocity is the speed tangential to the  trajectory

3) the data for this exercise are

     distance = (number of steps) (distance of one step)

         d₁ = 500 0.40 m = 200 m West

         d₂ = 400 0.35 m = 140 m North

         d₃ = 300 0.30 m = 90 m East

in a total time of t = 2 min and 120 s

           t = 2  60 + 120 = 240 s

the distance traveled is

x axis we assume the direction towards + x (East) positive

           x_total = d₃ -d₁

           x_total = 90-200 = -110 m

y axis  

          y_total = d₂

          y_total = 140 m

we use the Pythagorean theorem to find the distance0

           D = √ (x_total² + y_total²)

           D = √ (110² + 90²)

           D = 142.13 m

The speed is

           v = D / t

            v = 142.13 / 240

           v = 0.592 m / s

speed is

           v = 0592 m / s

           tan θ = Y / X

           θ = tan -1 y / x

           θ = tan -1 (90 / (- 110))

           θ = -39.3 º

4) This angle is measured on the positive side of the x axis in a clockwise direction

          θ = 1.5 rev (2π rad / 1 rev) = 3π rad

           t = 90 s

the angular velocity is

           w = θ / t

           w = 3πi / 90

           w = 0.1047 rad / s

the speed is

            v = w R

            v = 0.14047 250

            v = 26.17 m / s

velocity is tangent to path

Answer:

3m/s^2

Explanation:

Acceleration is change in velocity / time = 45/15 = 3m/s^2

Explanation:

satellite A, because of its closest orbit.

Answer:

Explanation:D

The velocity of the gold puck after the collision is equal to (2i - j) m/s. Therefore, option (A) is correct.

The rate at which a displacement of the body changes in relation to time is called its velocity. Velocity is a vector parameter with both magnitude and direction. S.I. unit of velocity can be represented as meter/second.

Given, the mass of the gold puck, m = 12 kg

The velocity of the gold puck before the collision, u =  5i – 4j m/s.

The mass of the stationary blue puck, M = 18 kg.

The velocity of it after the collision, V = 2i – 2j m/s.

We have to determine the velocity of the gold puck after the collision:

The total momentum of the system before the collision is equal to the total momentum of the system after the collision.

mu + M ×0 = mv + MV

v = u - MV/m

v  = (5i – 4j) m/s  - (2i – 2j)m/s × 18/12

v  = (5i – 4j) m/s - (3i - 3j)

v = (2i - j) m/s.

Therefore, the velocity of the gold puck is (2i - j) m/s after the collision.

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Answer:

As the projectile starts it’s motion from a vertical height at y=10m, it’s vertical motion may be described as under-

y= 10+ut-0.5gt^2

Where “u” is initial vertical velocity

= 45 sin60,

“g” is acceleration due to the gravity (10m/sec^2)

and “t” is time.

When the projectile touches the ground the vertical height “y”=0

Therefore the time “t” taken for the flight of the projectile till it touches the ground may be calculated from the above equation when y=0 as under -

0= 10+45sin60 x t- 0.5x10 t^2

0= 10+45x (√3/2)x t-5 t^2

t^2-9x (√3/2)x t-2 =0

t= [4.5√3 + √(243/4+4)]/2

t=[7.794+ 8.047]/2

t= 7.92 sec.

The horizontal distance travelled during this time=

vt

Where “v” is horizontal velocity = 45 cos60.

Therefore,

vt =45cos60 x 7.92

= 45x0.5x7.92

= 178.2 m

Answer:

The horizontal component of the velocity is 27.19 ft/s

The vertical component of the velocity at takeoff is 12.68 ft/s

The vertical distance traveled by the jumper is 8.2 ft

Explanation:

Given;

angle of projection or takeoff, θ = 25°

the resultant velocity, v = 30 ft/s

The horizontal component of the velocity is given by;

[tex]v_x = vcos \theta\\\\v_x = 30 * cos(25)\\\\v_x = 27.19 \ ft/s[/tex]

The vertical component of the velocity at takeoff is given by;

[tex]v_y = vsin \theta\\\\v_y = 30 * sin(25)\\\\v_y = 12.68 \ ft/s[/tex]

The vertical distance traveled by the jumper is given by;

[tex]v_y_f^2 = v_y_o^2 +2(-g)h\\\\v_y_f^2 = v_y_o^2 -2gh\\\\0 = 12.68^2 - (2*9.8)h\\\\0 = 160.78 - 19.6h\\\\19.6h = 160.78\\\\h = \frac{160.78}{19.6}\\\\ h = 8.2 \ ft[/tex]

Answer:

The correct option is;

3. layers of cotton lining the interior

Explanation:

In the procedure for making a solar oven it is essential place the inside portion of the carboard flap be made that the air made hot by the heat of the Sun does is not removed from the oven as convection current, and to provide insulation that limits heat loss by thermal radiation as well as conduction to cooler surfaces.

Answer:

^

Explanation:

Answer:

C) he had more power running than walking

Explanation:

saw it on a quizlet. not 100% sure tho

Hiro walks up a flight of stairs in his apartment building. Later in the day, he runs up the same flight of stairs so He had more power running than walking.

In physics, power is the amount of energy transferred or converted per unit of time. In the International System of Units, the unit of power is the watt, equal to one joule per second. In older works, power is sometimes called activity. Power is a scalar quantity.

Power is related to other quantities; for example, the power involved in moving a ground vehicle is the product of the traction force on the wheels and the velocity of the vehicle.

The output power of a motor is the product of the torque that the motor generates and the angular velocity of its output shaft

Since we know that Hiro will run on stairs to minimize the time of reaching his home so if time decreases then he has to increase his power as power is the ratio of work and time.

Hence Hiro walks up a flight of stairs in his apartment building. Later in the day, he runs up the same flight of stairs so He had more power running than walking.

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Answer:

Disliking something

Explanation:

For an example : Ah you disgust me

Answer:

Strongly disproval of something (usual this something is gross)

"Profound dislike or annoyance caused by something sickening or offensive." - The Free Dictionary

Explanation:

Example: Her attitude filled me with disgust

Example 2: Your gross remarks disgust me.

Answer:

mechanical energy

Explanation:

Answer: 3.75 m/s

Explanation:

Here's what you know: The car has mass 1500 kg, and initial velocity 15.0 m/s.  The truck has mass 4500 kg and initial velocity 0 m/s.  The final mass is 6000 kg (4500 kg plus 1500 kg, because they stick together) and we are looking for the final velocity.  Using the formula p = mv, and the concept that p before equals p after, the p of the car plus p of the truck equals the p of them together. So, 1500 x 15 + 4500 x 0 = 6000 x v final.  22500 = 6000 v, and v = 3.75 m/s.  The velocity is positive, and the initial velocity of the car was positive, so the two vehicles together are traveling to the south just like the car was initially.

Answer:

I believe it qould be A) The Force of gravity between two objects is directly related to the mass of the largest object.

Explanation:

Answer:

the answer to your question is b a mirror fogs up when someone takes a hot shower.

The process of conversion of gaseous water to its liquid form is called condensation. The fogs formed on the mirror from the hot shower is an example of condensation.

Condensation is the process of cooling up of water vapor to liquid water. Water vapor condenses when it cools. Condensation can be best understood from the reason behind raining.

When water vaporizers from resources and cools from the sky, and the vapor condenses to form liquid droplets. Similarly we can observe water droplets in the window pane due to the similar effect.

The water vapor arises from the hot shower condenses in the atmosphere and forms the drops on the mirror. Therefore, the mirror fogs up by the hot shower is an example of condensation.

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Answer:

ntensive properties: A physical property that will be the same regardless of the amount of matter.

Explanation:

Answer:

Explanation:

The force acting on an object can be found by using the formula

From the question

mass = 45 kg

acceleration = 1.5 m/s²

We have

Force = 45 × 1.5

We have the final answer as

Hope this helps you

Answer:

v = 0

Explanation:

Given that,

Total distance is 50 yards

Dugan got an early lead by finishing the first 25.00 yd in 10.01 seconds

Dugan finished the return leg (25.00 yd distance) in 10.22 seconds.

We need to find Dugan's average velocity for the entire race. As he returns at the initial position. As a result, the net displacement is equal to 0. So,

Average velocity = net displacement/time

v = 0

Hence, his average velocity for the entire race is 0.

Answer:

The acceleration is 2 m/s² in Camila's direction.

Explanation:

The acceleration can be found using the following equation:

[tex] \Sigma F = ma [/tex]

Where

F is the force

m is the mass of the taffy = 1.5 kg

a is the acceleration =?

Since Emma pulls one end of the taffy with 20 N and Camila pulls the other end with 23 N we have:

[tex] F_{C} - F_{E} = ma [/tex]

[tex]23 N - 20 N = 1.5 kg*a [/tex]

[tex] a = 2 m/s^{2} [/tex]

Hence, the acceleration is 2 m/s² in Camila's direction.

I hope it helps you!

Answer:

Out of this, Area is not a fundamental physical quantity.

Answer:

10m/s^2

Explanation:

a=change in velocity/change in time

50/5=10m/s^2

Given :

Initial velocity, u = 12.5 m/s.

Height of camera, h = 64.3 m.

Acceleration due to gravity, g = 9.8 m/s².

To Find :

How long does it take the camera to reach the ground.

Solution :

By equation of motion :

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

Putting all given values, we get :

[tex]12.5t+\dfrac{9.8t^2}{2}=64.3\\\\4.9t^2+12.5t=64.3[/tex]

t = 2.56  and t = −5.116.

Since, time cannot be negative.

t = 2.56 s.

Therefore, time taken is 2.56 s.

Hence, this is the required solution.

Answer: .4 m/s^2= acceleration

Explanation:

f = m*a

We can rearrange this equation to solve for acceleration. Therefore,

a=f/m

a= 28N/70kg

a= 0.4 m/s^2

The answer is 388 800 sec

Answer:

388800 secs

Explanation:

1 day = 24 hrs

so 24 times 3 = 72

and 1 hour is equal to 60 min

108 x 60 = 6480

and 60 secs in a minute

6480 x 60= 388800

Answer:

The distance to come to a halt is approximately 49.53 meters

Explanation:

Thee given parameters are;

The speed of the car, v = 50 mph = 22.35 m/s

The mass of the car, m = 1000 kg

The coefficient of friction,  = 0.514

The force of friction of the brake = Mass × Gravity × Friction = 1000 × 9.81 × 0.514 = 5042.34 N

The initial kinetic energy of the car = 1/2×m×v² = 1/2 × 1000 × 22.35² = 249761.25 J

The work done by the brake = Force of the brake × Distance, d, to come to halt

By conservation of energy, we have;

The work done by the brake = The initial kinetic energy of the car

∴ The initial kinetic energy of the car = Force of the brake × Distance, d, to come to halt

The initial kinetic energy of the car = 249761.25 J = 5042.34 N × Distance, d, to come to halt

∴The distance to come to a halt = 249761.25 J /(5042.34 N) ≈ 49.53 meters

The distance to come to a halt ≈ 49.53 meters.

The car will cover 49.65 m distance before stopping due to application of brake.

Given data:

The mass of car is, M = 1000 kg.

The initial speed of car is, u = 50.0 mph = (50)(0.447 m/s) = 22.35 m/s.

The coefficient of static friction is, [tex]\mu = 0.514[/tex].

The given problem can be solved using the third kinematic equation of motion. Which is,

[tex]v^{2}=u^{2}+2as[/tex] ....................................................(1)

Here, v is the final speed, v = 0 (Because car is finally stopping)

s is the distance covered before stopping and a is the magnitude of acceleration.

Now, since frictional force opposes the motion. Then,

[tex]F = f\\\\ma = \mu \times m \times g\\\\a = 0.514 \times 9.8\\\\a =5.03 \;\rm m/s^{2}[/tex]

Substituting the values in equation (1) as,

[tex]0^{2}=23.35^{2}+2(-5.03)s\\\\s = \dfrac{2500}{2 \times 5.03}\\\\ s=49.65 \;\rm m[/tex]

Thus, the car will cover 49.65 m distance before stopping due to application of brake.

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