PHYSICS COLLEGE

What is the magnitude of a vector that has the following components: x = 32 m y = -59 m

Answers

Answer 1

Answer:

Explanation:

Since the x and y components are given

The vectors Magnitude = √32²+(-59)²

=67.12m

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Assume the electric field has a magnitude of 59.0 m V , and the particle has a net positive charge of Q=+5.40 C. Calculate the magnitude of the electric potential difference, \vert \Delta V \vert = \vert V_B - V_A \vert∣ΔV∣=∣V B −V A ∣.

Answers

Answer:

V = 1.69 * 10^6 V

Explanation:

Parameters given:

Electric field, E = 59V/m

Charge, q = 5.40C

We need to first find the distance between the electric charge and the point of consideration to be able to find the Electric potential difference.

Electric field is given as:

E = (kq/r^2)

k = Coulombs constant

=> r^2 = kq/E

=> r^2 = (9 * 10^9 * 5.4) / 59

r^2 = 8.2 * 10^8

r = 2.84 * 10^4 m

We can now find the Electric Potential by using:

V = kq/r

Hence,

V = (9 * 10^9 * 5.4) / (2.84 * 10^4)

V = 1.69 * 10^6 V

A 100-N uniform ladder, 8.0 m long, rests against a smooth vertical wall. The coefficient of static friction between ladder and floor is 0.40. What minimum angle can the ladder make with the floor before it slips?

Answers

The minimum angle that the ladder make with the floor before it slips is 51.34 Degree.

Given data:

The weight of ladder is, W = 100 N.

The length of ladder is, L = 8.0 m.

The coefficient of static friction between ladder and floor is, $\mu =0.40$ .

Apply the Newton' law in vertical direction to obtain the value of Normal Force (P) as,

$N = mg$

And force along the horizontal direction is,

$F= \mu * N\n\nF = \mu * mg$

Now, taking the torque along the either end of ladder as,

$-mgcos \theta * (L)/(2)+Fsin \theta * L =0\n\nmgcos \theta * (L)/(2) = Fsin \theta * L$

Solving as,

$mgcos \theta * (L)/(2) = (\mu mg) * sin \theta * L\n\ncos \theta * (1)/(2) = (\mu) * sin \theta\n\ntan \theta = (1)/(2 * 0.40 )\n\n\theta = tan^(-1)(1/0.80)\n\n\theta = 51.34^(\circ)$

Thus, we can conclude that the minimum angle that the ladder make with the floor before it slips is 51.34 Degree.

Learn more about the frictional force here:

brainly.com/question/4230804

Answer:

The minimum angle is 51.34°

Explanation:

Given that,

Weight of ladder = 100 N

Length = 8.0 m

Coefficient of static friction = 0.40

We need to calculate the normal force

Using Newtons law in vertical direction

$F_(y)=n-mg$

$N-mg=0$

$N=mg$

We need to calculate the normal force

Using Newtons law in horizontal direction

$F_(s)=f_(s)-P$

$f_(s)-P=0$

$f_(s)=P$

$P=\mu mg$

We need to calculate the minimum angle

Using torque about the point A then

$-mg\cos\theta* AB+P\sin\theta* AC=0$

Put the value into the formula

$mg\cos\theta*((L)/(2))=\mu mg\sin\theta* L$

$\cos\theta*(1)/(2)=\mu\sin\theta$

$(1)/(2)=\mu*\tan\theta$

$\theta=\tan^(-1)((1)/(2*0.40))$

$\theta=51.34^(\circ)$

Hence, The minimum angle is 51.34°

If a pressure gauge measure an increase in 3×10^(5)Pa on an area of 0.7 m^2, calculate the increase in the force applied to the area?

Answers

Answer:210000N

Explanation:

Pressure=3x10^5pa

area=0.7m^2

Force = pressure x area

Force=3x10^5x0.7

Force=210000N

A boy throws a snowball straight up in the air with an initial speed of 4.50 ft/s from a position 4.00 ft above the ground. The snowball falls straight back down in to a 6 inches of snow. The snowball feels a deceleration of 100 m/s2 upon impact with the snow bank before coming to rest. (a) When does the snowball hit the top of the snow bank? (b) How far from the ground does the snowball come a rest?

Answers

Answer:

a) 0.658 seconds

b) 0.96 inches

Explanation:

$v=u+at\n\Rightarrow 0=4.5-32.1* t\n\Rightarrow (-4.5)/(-32.1)=t\n\Rightarrow t=0.14 \s$

Time taken by the ball to reach the highest point is 0.14 seconds

$s=ut+(1)/(2)at^2\n\Rightarrow s=4.5* 0.14+(1)/(2)* -32.1* 0.14^2\n\Rightarrow s=0.315\ ft$

The highest point reached by the snowball above its release point is 0.315 ft

Total height the snowball will fall is 4+0.315 = 4.315 ft

$s=ut+(1)/(2)at^2\n\Rightarrow 4.315=0t+(1)/(2)* 32.1* t^2\n\Rightarrow t=\sqrt{(4.315* 2)/(32.1)}\n\Rightarrow t=0.518\ s$

The snowball will reach the bank at 0.14+0.518 = 0.658 seconds after it has been thrown

$v=u+at\n\Rightarrow v=0+32.1* 0.518\n\Rightarrow v=16.62\ ft/s$

$v^2-u^2=2as\n\Rightarrow s=(v^2-u^2)/(2a)\n\Rightarrow s=(0^2-16.62^2)/(2* -100* 3.28)\n\Rightarrow s=0.42\ ft$

The snowball goes 0.5-0.42 = 0.08 ft = 0.96 inches

a trcuk weighs four times as much as a stationary car. if teh truck coasts into the car at 12 km/s and they stick toegther, what is their final velocity

Answers

Answer:

v=9.6 km/s

Explanation:

Given that

The mass of the car = m

The mass of the truck = 4 m

The velocity of the truck ,u= 12 km/s

The final velocity when they stick = v

If there is no any external force on the system then the total linear momentum of the system will be conserve.

Pi = Pf

m x 0 + 4 m x 12 = (m + 4 m) x v

0 + 48 m = 5 m v

5 v = 48

$v=(48)/(5)\ km/s$

v=9.6 km/s

Therefore the final velocity will be 9.6 km/s.

An insulated rigid tank contains 3 kg of H2O in the form of a saturated mixture of liquid and vapor at a pressure of 150 kPa and a quality of 0.25. An electric heater inside the tank is turned on to heat this H2O until the pressure increases to 200 kPa. Please determine the change in total entropy of water during this process. Hint: See if you can find the electrical work consumed during this process.