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3 years ago

What does it mean to control variables

1 answer:

3 0

Controlled variables are influences that could affect the outcome of an experiment, and so are purposely controlled so that they do not impact the experimental results.

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The diagram below shows a golf ball being struck by a club. The ball leaves the club with a speed of 40 meters per second at an

hjlf

Answer:

61.22m

Explanation:

Maximum height in projectile is expressed as;

H = u²sin²θ/2g

u is the speed

g is the acceleration due to gravity

θ is the angle of projection

Substitute the given values into the formula;

H = u²sin²θ/2g

H = 40²sin²60/2g

H = 1600(0.8660)²/2(9.8)

H = 1,199.9/19.6

H = 61.22m

Hence the maximum height achieved by the ball is 61.22m

4 0

3 years ago

A 3,220 lb car enters an S-curve at A with a speed of 60 mi/hr with brakes applied to reduce the speed to 45 mi/hr at a uniform

Grace [21]

The magnitude of the total friction force exerted by the road on the tires at B will be 919.46 lb.

<h3>What is the friction force?</h3>

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. Its unit is Newton (N).

Mathematically, it is defined as the product of the coefficient of friction and normal reaction.

The given data in the problem is;

The weight is,W= 3,220 lb

The speed is,u= 60 mi/hr

The reducing speed is,v= 45 mi/hr

The distance traveled is,d= 300 ft

The radius of curvature of the path of the car at B is,R= 600 ft.

1 mile = 5280 ft

From the Newtons' equation of motion;

$\rm v^2 = u^2 +2ad \\\\ \rm (45 \times \frac{5280}{3600} )^2 = (60 \times \frac{5280}{3600} )^2 +2a\times 300 \\\\$

The tangential accelerations are;

$\rm a_t = \frac{66^2 -88^2}{600} \\\\ \rm a_t = -5.65 ft/sec^2 \\\\$

The force is found as;

$\rm \sum F = ma \\\\ \frac{3220}{32.2} \times -5.65 \\\\ F_T= 565 \ lb$

The normal force is;

$\rm F_n = \frac{3220}{32.2} \times \frac{66^2}{600} \\\\ F_N =726 \ lb$

The net or the total friction force exerted by the road on the tires at B. is found as;

$\rm F = \sqrt{(565)^2+(726)^2} \\\ F = 919.946 \ lb$

Hence, the magnitude of the total friction force exerted by the road on the tires at B will be 919.46 lb.

To learn more about the friction force, refer to the link;

brainly.com/question/1714663

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6 0

2 years ago

(a) The stick is supported by a sharp point at the middle. On the left side, a weight of 100 g is suspended at 40 cm from the mi

Jet001 [13]

Answer:

20cm

Explanation:

Hello!

remember that the condition for a body to be at rest is that the sum of its moments and its forces be zero,

To solve this problem you must draw the free body diagram of the stick (attached image) and sum up moments at point 0 (where the sharp is located), which results in the following equation

(100g)(40cm)=x(200g)

$X=\frac{(100)(40)}{(200)} =20cm$

6 0

3 years ago

Two atoms with different mass number but the same atomic number are called A) elements. B) ions. C) isotopes. D) nuclei.

saw5 [17]

C isotopes is the correct answer. Isotopes are variants of a particular chemical element which differ in neutron number, and consequently in nucleon number.All isotopes of a given element have the same number of protons but different numbers of neutrons in each atom.

Hope this helps! :)

4 0

3 years ago

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Location C is 0.02 m from a small sphere which has a charge of 3 nanocoulombs uniformly distributed on its surface. Location D i

kkurt [141]

The change in potential along a path from C to D due to a small charged sphere is 900 V.

Given:

Charge, Q = 3 nC = 3 × 10⁻⁹ C

Distance between the sphere and point C, r₁ = 0.02 m

Distance between the sphere and point D, r₂ = 0.06 m

Calculation:

We know that the electric potential is given as:

V = k Q/r - (1)

where, V is the electric potential

k is Coulomb's force constant

Qis the charge on the sphere

r is the separation distance

The electric potential at point C due to charged sphere can be given as:

V₁ = k Q/r₁

= (9×10⁹ Nm²/C²) [(3 × 10⁻⁹ C)/(0.02 m)]

= 1350 V

The electric potential at point D due to charged sphere can be given as:

V₂ = k Q/r₂

= (9×10⁹ Nm²/C²) [(3 × 10⁻⁹ C)/(0.06 m)]

= 450 V

Now, the change in potential along the path from C to D can be calculated as:

ΔV = V₂ - V₁

= 450 V - 1350 V

= -900 V

The negative sign indicates that the work is done against the electric field in moving the charge from C to D.

Therefore, the change in potential along a path from C to D is 900 V against the direction of the electric field.

Learn more about the electric potential here:

<u>brainly.com/question/12645463</u>

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8 0

1 year ago

Read 2 more answers