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3 years ago

Circular motion requires the application of a constant force in which direction?

1 answer:

6 0

Toward the center of the circle

The final motion characteristic for an object undergoing uniform circular motion is the net force. The net force acting upon such an object is directed towards the center of the circle. The net force is said to be an inward or centripetal force.

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Suppose a car is traveling in the negative x-direction and comes to a stop. What is the sign of that car’s acceleration? How do

Oduvanchick [21]

We have equation of motion v = u + at

Where v = final velocity, u = initial velocity, a = acceleration, and t = time

In this case car is travelling in -x direction

Velocity of car = displacement / time

Since displacement value is increasing in negative x axis it's initial velocity is negative

And it's final velocity is zero since it comes to rest, and time is also positive

So, v= u+ at => 0 = -u + at

So, a = u/t

Which is positive and along positive X - direction

7 0

4 years ago

g A particle (charge = +40 mC) is located on the x axis at the point x = -20 cm, and a second particle (charge = -50 mC) is plac

krok68 [10]

Answer:

Explanation:

We shall find electric field at origin due to two given charges sitting on the either side of origin .

Total field will add up due to their same direction .

Field due to a charge Q

= 9 x 10⁹ x Q / R² ; R is distance of point , Q is charge

Field due to first charge

= 9 x 10⁹ x 40 x 10⁻³ / 2² x 10⁻⁴

= 90 x 10¹⁰ N/C

Field due to second charge

= 9 x 10⁹ x 50 x 10⁻³ / 2² x 10⁻⁴

= 112.5 x 10¹⁰ N/C

Total field

= 202.5 x 10¹⁰ N/C

Force on given charge at origin

= charge x field

= 4 x 10⁻³ x 202.5 x 10¹⁰

= 810 x 10⁷ N .

4 0

4 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

Two pulses are moving along a string. One pulse is

raketka [301]

Answer:

The answer is the 3rd option!

6 0

3 years ago

A windmill is example of a <br>a.generator <br>b.resistor <br>c.insulator<br>d.motor

Alexandra [31]

The answer is a. generator!

4 0

3 years ago