An increase in the magnitude of which others quantities causes a reduction in centripetal force?
A. Mass
B.Radius of curvature
C. Velocity

Answer:

6km

Explanation:

Answer:

The final velocity of the 2kg ball is 1.270 m/s

Explanation:

According to Newton's second and third laws of motion

Newton's second law state that "the rate of change of momentum is proportional to the applied force and takes place in the direction of that force".

Newton's third law state that "for every action, there must be an equal and opposite reaction".

The combinations of these two laws resulted in an elastic collision

Given that:

m1 = 2kg

u1 = 2.20m/s

m2 = 4.00kg

u2 = 0m/s

An Elastic collision is when kinetic energy before = kinetic energy after

E.K before = [tex]1/2mv^{2}[/tex]

E.K before = 1/2 * 2 * (2.20)^2

E.K = 1/2 * 2 * 4.84

E.K before = 4.84j

E.K after = 1/2 x (4 + 2)v^2

E.K after = 1/2(6v^2)

E.K after = 3v^2

Since E.K before = E.K after

4.84 = 3v^2

Divide through by 3

4.84/3 = 3v^2/3

1.6133 = v^2

[tex]V = \sqrt{1.6133} \\V = 1.270 m/s[/tex]

Answer:

d

Explanation:

Answer:

B. Randomly assigning students to the two different groups

Explanation:

The bias can be controlled by making use of the randomized block design where we divide the subject into subgroups also called blocks and randomly assign treatment conditions to each block.

In this case, the blocks or sub groups are learning math from a video game and learning it from traditional classroom activities.

So the best way to control the bias is by randomly assigning students to the different groups.

Answer:

Explanation:

Thermal energy or internal energy gain or loss = mass x specific heat x temperature  

specific heat of water = 4.2 kJ / kg degree Celsius

mass of water lost in 10 second = rate of loss x time = .5 x 10 = 5 kg .

heat energy associated with lost water = 5 x 4.2 x ( 60 - 0 ) = 1260 kJ .

Heat energy lost = 1260 kJ .

Answer:

1.08 m/s^2

Explanation:

The maximum value of static friction F_s = μ_s mg

where μ_s = coefficient of static friction

m= mass of box= 50kg

g= 9.81 m/s^2 , acceleration due to gravity

F_s = 0.5×50×9.81 =245 N

Since, the applied force is greater than the maximum value of static friction F_s, the object must be in motion. Hence, kinetic friction must be taken into account and not static friction.

By equilibrium condition

ma = F - μ_k m g

where F = 250 N,  μ_k = coefficient of kinetic friction = 0.4, m= 50 Kg

a = acceleration of the box

50a = 250-0.4×50×9.81

Solving we get a=  1.08 m/s^2

Answer: 1

Explanation:

Given

Tension is the string [tex]T=44\ N[/tex]

mass of object [tex]m=4\ kg[/tex]

Tension is greater than the weight of the object i.e. elevator is moving upward

we can write

[tex]\Rightarrow T-mg=ma\\\Rightarrow T=m(g+a)\\\Rightarrow 44=4(10+a)\\\Rightarrow 11=10+a\\\Rightarrow a=1\ m/s^2[/tex]

Answer:

Cayla ? please whats going on ?

Explanation:

Answer:

Height of the ramp

Explanation:

The independent variable is always what the person doing the experiment can change or modify.  So if the question is about whether the height of the ramp with  effect the car, that is indeed what they are changing or modifying in the experiment.

Answer:

Static friction is what keeps the box from moving without being pushed, and it must be overcome with a sufficient opposing force before the box will move. Kinetic friction is the force that resists the relative movement of the surfaces once they're in motion.

Answer:

conduction (the heat is transferring to the air)

Answer and Explanation: Momentum (Q) is defined as mass in motion, which means it relates mass of an object with its velocity:

Q = m.v

So, momentum only depends on mass of the object and the velocity it is moving.

Comparing bowling ball and ping pong ball, the first one has more mass than the second one. Therefore, a bowling pin will better knock down the pins than the ping pong ball.

Answer:

the laws of physics are the same for all non-accelerating observers

Explanation:

Answer:

its the third one

Answer:

354 m/s

Explanation:

For the second overtune (Third harmonic) of an open pipe,

λ = 2L/3................................ Equation 1

Where L = Length of the open pipe, λ = Wave length.

Given: L = 1.75 m.

Substitute into equation 1

λ = 2(1.75)/3

λ = 1.17 m.

From the question,

V = λf.......................... Equation 2

V = speed of sound in the room, f = frequency

Given: f = 303 Hz.

Substitute into equation 2

V = 1.17(303)

V = 353.5

V ≈ 354 m/s

Hence the right answer is 354 m/s

The satement that can be made about sound wave A and sound wave B is, they will slow down.

The relationship between sound waves and temperature is given by the following formula;

[tex]v = \sqrt{\frac{\gamma RT}{M} }[/tex]

The speed of sound wave increases with increase in temperature, and vice versa.

Sound wave is mechanical wave, because it requires material medium for its propagation. Sound will travel faster in liquid medium than gaseous medium because of number of molecules per unit volume.

Thus, the satement that can be made about sound wave A and sound wave B is, they will slow down.

Learn more about speed of sound waves here: https://brainly.com/question/2142871

Answer:

Height of its bounce

Explanation:

The dependent variable is always what is being measured or the data collected.

q1 = -5.00 x 10-6 C

q2 = -5.00 x 10-6 C

q3 = -5.00 x 10-6 C

E1 = kq/r^2 = ( 9 x 10^9)( 5 x 10^-6)/(0.5^2) = 180000 N/C to the left

E2 = kq/r^2 = ( 9 x 10^9)( 5 x 10^-6)/(0.25^2) = 720000 N/C to the right

E net = 720000 - 180000 = 540000 N/C to the right

F = qE

F = (-5 x 10^6 C)(540000 N/C) = - 2.7 N

The force on q2 is 2.7 N to the left.

The net electrostatic force on the q2 is 2.7N owards left

The equation for electrostatic force is

        [tex]F= k\frac{q_{1}q_{2} }{r^{2} }[/tex]

where k = [tex]9*10^{9} Nm^{2}/C^{2}[/tex] and r is the distance separating charges q1 and q2.

the force has to be calculated on a charge q2 = -5.0 ×[tex]10^{-6}[/tex] C by the charges q1=  -5.0 ×[tex]10^{-6}[/tex] C and q3=  -5.0 ×[tex]10^{-6}[/tex] C

distance between q1 and q2 is 0.5 m = 5×[tex]10^{-1}[/tex]m

distance between q2 and q3 is 0.25 m = 25×[tex]10^{-2}[/tex]m

force due to charge q1

           [tex]F_{1}[/tex] = 9×[tex]10^{9}[/tex]×(-5)×(-5)×[tex]10^{-12}[/tex]/25×[tex]10^{-2}[/tex] N = +0.9N =  0.9N towards right

           [tex]F_{2}[/tex] = 9×[tex]10^{9}[/tex]×(-50)×(-4)×[tex]10^{-12}[/tex]/625×[tex]10^{-4}[/tex] N = -3.6N = 3.6N towards left

hence net force F = [tex]F_{1}+F_{2}[/tex]

                              = 0.9N - 3.6N = -2.7N

                           F = 2.7 N towards left

Learn more about electrostatic force:

https://brainly.com/question/11375177

Answer:

a. i. 0.044 ft³/s ii. 2.75 lb/s

b. 62.3 s

c. 32 ft/s

Explanation:

(a) the volume and mass flow rates of water through the hose

i. The volume flow rate

The volume flow rate, Q = Av where A = cross-sectional area of nozzle at inlet = πd²/4 where r = diameter of nozzle at inlet = 1 inch = 1/12 ft and v = average velocity in the nozzle = 8 ft/s.

So, Q = Av

= πd²v/4

= π(1/12 ft)² × 8 ft/s/4

= 0.175/4 ft³/s

= 0.044 ft³/s

ii. The mass flow rate

The mass flow rate, M = ρQ where ρ = density of water = 62.4 lb/ft³ and Q = volume flow rate = 0.044 ft³/s

So, M = ρQ

= 62.4 lb/ft³ × 0.044 ft³/s

=  2.745 lb/s

≅ 2.75 lb/s

(b) how long it will take to fill the bucket with water

Since we have 20 gallons = 20 × 1 gallon, we convert if to cubic feet. So, 20 gallons = 20 × 1 gallon = 20 × 0.137 ft³ = 2.74 ft³.

Since V = Qt where V = volume, Q = volume flow rate and t = time, making t subject of the formula, we have

t = V/Q

= 2.74 ft³/0.044 ft³/s

= 62.27 s

≅ 62.3 s

(c) the average velocity of water at the nozzle exit

Using the continuity equation,

A₁v₁ = A₂v₂ where A₁ = cross-sectional area of nozzle at inlet = πd₁²/4 where d₁ = diameter of nozzle at inlet = 1 inch = 1/12 ft, v₁ = average velocity in the nozzle = 8 ft/s, A₂ = cross-sectional area of nozzle at inlet = πd₂²/4 where d₂ = diameter of nozzle at inlet = 0.5 inch = 0.5/12 ft, v₂ = average velocity at exit nozzle.

So, A₁v₁ = A₂v₂

(πd₁²/4)v₁ = (πd₂²/4)v₂

πd₁²v₁/4 = πd₂²v₂/4

Dividing through by π/4. we have

So, d₁²v₁ = d₂²v₂

making v₂ subject of the formula, we have

v₂ = d₁²v₁/d₂²

v₂ = (d₁²/d₂²)v₁

v₂ = (d₁/d₂)²v₁

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

v₂ = (1/0.5)² × 8 ft/s

v₂ = (2)² × 8 ft/s

v₂ = 4 × 8 ft/s

v₂ = 32 ft/s

Answer:

terminal velocity is;

v = 117.54 m/s

v = 423.144 km/hr

Explanation:

Given the data in the question;

we know that, the force on a body due to gravity is;

[tex]F_g[/tex] = mg

where m is mass and g is acceleration due to gravity

Force of drag is;

[tex]F_d[/tex] = [tex]\frac{1}{2}[/tex]pCAv²

where p is the density of fluid, C is the drag coefficient, A is the area and v is the terminal velocity.

Terminal velocity is reach when the force of gravity is equal to the force of drag.

[tex]F_g = F_d[/tex]

mg =  [tex]\frac{1}{2}[/tex]pCAv²

we solve for v

v = √( 2mg / pCA )

so we substitute in our values

v = √( [2×(86 kg)×9.8 m/s² ] / [ 1.21 kg/m³ × 0.7 × 0.145 m²] )

v = √( 1685.6 / 0.122015 )

v = √( 13814.6949 )

v  = 117.54 m/s

v = ( 117.54 m/s × 3.6 ) = 423.144 km/hr

Therefore terminal velocity is;

v = 117.54 m/s

v = 423.144 km/hr

Explanation:

angular velocity = velocity/radius

= 5/2

= 2.5 rad/s

Positive

In an LED, the longer lead is always connected to the positive terminal 

Answer:

Aerobics I think.

Explanation:

Answer:

Explanation:

Force of friction at car B ( break was applied by car B ) =μ mg = .65 x  2100 X 9.8  = 13377 N .

work done by friction = 13377 x 7.30 = 97652.1 J

If v be the common velocity of both the cars after collision

kinetic energy of both the cars = 1/2 ( 2100 + 1500 ) x v²

= 1800 v²

so , applying work - energy theory ,

1800 v² = 97652.1

v² = 54.25

v = 7.365 m /s

This is the common velocity of both the cars .

To know the speed of car A , we shall apply law of conservation of momentum  .Let the speed of car A before collision be v₁ .

So , momentum before collision = momentum after collision of both the cars

1500 x v₁ = ( 1500 + 2100 ) x 7.365

v₁ = 17.676 m /s

= 63.63 mph .

( b )

yes Car A was crossing speed limit by a difference of

63.63 - 35 = 28.63 mph.

(a) The speed of car A just before the collision is 51.58 mph.

(b) With the given speed limit of 35 miles per hour, car A was crossing the speed limit by 16.58 mph.

What is collision?

The event when two objects strike each other from either direction, then such event is known as a collision. During the collision, the speed of colliding objects may vary according to the direction of the approach.

Given data -

The mass of car A is,  mA = 1500 kg.

The mass of car B is, mB = 2100 kg.

The length of the skid mark is, d = 7.30 m.

The coefficient of kinetic friction between tires and road is, [tex]\mu = 0.65[/tex].

(a)

The combined kinetic energy of both cars is,

[tex]KE_{T}=\dfrac{1}{2} (mA+mB)v^{2}\\\\KE_{T}=\dfrac{1}{2} (1500+2100)v^{2}\\\\KE_{T}=1800v^{2}[/tex]

Applying the work-energy principle as,

Work done due to kinetic friction = Combined kinetic energy of cars

[tex]F \times d = KE_{T}\\\\(\mu \times (mA+mB)\times g) \times d = KE_{T}\\\\(0.65 \times (1500+2100)\times 9.8) \times 7.30 = 1800v^{2}\\\\v = 9.64 \;\rm m/s[/tex]

Converting into mph as,

[tex]v = 9.64 \times 2.23\\\\v = 21.49 \;\rm mph[/tex]

To know the speed of car A , we shall apply the law of conservation of momentum. Let the speed of car A before collision be v₁.

So , momentum before collision = momentum after collision of both the cars

1500 x v₁ = ( 1500 + 2100 ) x 21.49

v₁ = 51.58 mph

Thus, we can conclude that the speed of car A just before the collision is 51.58 mph.

(b)

With the given speed limit of 35 mph, the obtained speed of car A before the collision is 51.58 mph. Clearly, car A is crossing the speed limit. And the difference is,

= 51.58 - 35 = 28.63 mph.

= 16.58 mph

Thus, we can conclude that car A was crossing the speed limit by 16.58 mph.

Learn more about the average speed here:

https://brainly.com/question/12322912

Explanation:

What impulse occurs when a cart that is originally at rest experiences an average force of

N for 2.5 s? *

(10 Points)

25 N

25 Nm

25 Ns

25 kg m/s

Answer:

Bitter Magnet inside a superconducting magnet

Explanation:

Since there are no options available, generally, the electromagnet that is considered the strongest is the Bitter Magnet inside a superconducting magnet.

This electromagnet produces 45 Tesla units which is a result of bitter magnet producing 33.5 Tesla and the superconducting coil produces the additional 11.5 Tesla.

Hence, justifying that the greater the current in the coil the stronger the electromagnet. ​