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4 years ago

The way an object moves relative to a reference point describes the object's _____.

Physics

1 answer:

Oduvanchick [21]4 years ago

6 0

Answer:

a measure of both length and the direction of an object's path from its starting point straight to its ending point. ... Distance between reference points and other locations can be measured in ... It Has Experienced Motion. When an object changes position over time, speed describes how quickly an object changes position.

Explanation:

You might be interested in

A loudspeaker has a circular opening with a radius of 0.1158 m. The electrical power needed to operate the speaker is 27.0 W. Th

Vilka [71]

Answer:

electrical power is converted by the speaker into sound power 17.25 %

Explanation:

given data

radius = 0.1158 m

power needed = 27.0 W

average sound intensity = 25.6 W/m²

solution

we know sound intensity that is

I = $\frac{power}{area}$ ...................1

power = intensity × area

Power = 25.6 × 4 ×π×r²

Power = 25.6 × 4 ×π×0.1158²

power = 4.313 W

so here electric power % into sound power

= $\frac{4.313}{25}$ × 100

= 17.25 %

3 0

3 years ago

Validity of an experiment is based upon

igor_vitrenko [27]

Validity concludes the whole experimental values and concepts which makes that whether the results which are obtained meet all of the requirements of the scientific research method.

On which factors the validity of an experiment depends?

There are five components which are used in experiment: observations, questions, hypothesis, methods and results. The scientific method procedure is not only ensures that the same experiment can be repeated by other researchers, but this experiment is accepted. For example, there must have been arrange in a random way of the sample groups and proper work and care shows and the experiment is succeed.

So we can conclude that: Validity concludes the whole experimental values and concepts which makes that whether the results which are obtained meet all of the requirements of the scientific research method.

Learn more about Validity: brainly.com/question/25303029

SPJ1

4 0

2 years ago

Two small plastic spheres each have a mass of 1.1 g and a charge of -50.0 nC . They are placed 2.1 cm apart (center to center).

Solnce55 [7]

Answer:

Part a)

F = 0.051 N

Part b)

Both spheres will follow Newton's III law of action reaction force so both sphere will have same force of equal magnitude.

Explanation:

Part a)

Electrostatic force between two charged spherical balls is given as

$F = \frac{kq_1q_2}{r^2}$

here we will have

$q_1 = q_2 = 50 nC$

here the distance between the center of two balls is given as

$r = 2.1 cm = 0.021 m$

now we will have

$F = \frac{(9\times 10^9)(50 \times 10^{-9})(50 \times 10^{-9})}{0.021^2}$

$F = 0.051 N$

Part b)

Both spheres will follow Newton's III law of action reaction force so both sphere will have same force of equal magnitude.

3 0

3 years ago

The concrete post (Ec = 3.6 × 106 psi and αc = 5.5 × 10-6/°F) is reinforced with six steel bars, each of 78-in. diameter (Es = 2

masha68 [24]

Answer:

the normal stress induced by the concrete post $\sigma_c = 67.26 psi$

the normal stress induced by the steel $\sigma_s = - 1795.84 psi$

Explanation:

Given that:

Modulus for elasticity for concrete post $E_c$ = $3.6 *10^6$ psi

Thermal coefficient for concrete post $\alpha _c$ = $5.5 *10^{-6}/^0F$

Modulus for elasticity of steel bar $E_s$ = $29*10^6psi$

Thermal coefficient of steel bar $\alpha _2$ = $6.5*10^{-6}/^0F$

Change in temperature ΔT = 80°F

Diameter of the steel rood = 7/8-in

Area of the steel rod $A_s$ = $6(\frac{ \pi}{4} )(d_s)^2$

= $6(\frac{ \pi}{4} )(\frac{7}{8} )^2$

= 3.61 in²

Area of concrete parts $A_c$ = (10)(10) - $A_s$

= (100 - 3.61) in²

= 96.39 in²

The total strain developed in the concrete post can be expressed as:

= $[\frac{1}{E_cA_c}+\frac{1}{E_sA_s}]P=(\alpha_s-\alpha_c)(\delta T)$

= $[\frac{1}{(3.6*10^6)(96.39)}+\frac{1}{(29*10^6)(3.61)}]P=(6.5*10^{-6}-5.5*10^{-6})(80)$

= $[(2.88*10^{-9}) +(9.55*10^{-9}]P = 8.0*10^{-5}$

= $1.234*10^{-8}P = 8.0*10^{-5}$

$P = \frac {8.0*10^{-5}}{1.234*10^{-8}}$

P = 6482.98 lb

Since, the normal stress in concrete is induced as a result of temperature rise; we have the expression :

$\sigma_c =\frac{P}{A_c}$

$\sigma_c = \frac{6482.98}{96.39}$

$\sigma_c = 67.26 psi$

Thus, the normal stress induced by the concrete post $\sigma_c = 67.26 psi$

Also; the normal stress in the steel bars induced as a result of temperature rise is as follows:

$\sigma_s = \frac{-P}{A_s}$

$\sigma_s =\frac{-6482.98}{3.61}$

$\sigma_s = - 1795.84 psi$

Thus, the normal stress induced by the steel $\sigma_s = - 1795.84 psi$

6 0

4 years ago

An object is pulled with a 85 N force inclined at 27° along a horizontal surface ABC

OLEGan [10]

Answer:

m = 15.15 kg

Explanation:

Newton's Second Law of motion states that when an unbalanced force is applied on a body, an acceleration is produced in it in the direction of force. The component of force along the horizontal direction here, will be given by the Newton's Second Law as:

Fx = ma

F Cosθ = ma

where,

F = Magnitude of Force = 85 N

θ = Angle with horizontal = 27°

m = mass of object = ?

a = acceleration of object = 5 m/s²

Therefore,

85 N Cos 27° = m(5 m/s²)

m = 75.73 N/5 m/s²

6 0

3 years ago