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77julia77 [94]
3 years ago
14

The force F has a magnitude of 500 lb and acts along the line AM where M is the midpoint of the vertical side OB of the parallel

epiped. Express F as its magnitude times the appropriate unit vector and then determine its x-, y-, and z-scalar components.

Physics
1 answer:
Sedbober [7]3 years ago
6 0

Answer:

The magnitude of components vector F (x,y,z) respectively =

390lb, -244lb, 195.2lb respectively

Explanation:

see the attached file

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I am sitting on a train car traveling horizontally at a constant speed of 50 m/s. I throw a ball straight up into the air.
drek231 [11]

Answer:

Explanation:

I am sitting on a train car traveling horizontally at a constant speed of 50 m/s. I throw a ball straight up into the air. Before , the ball gets separated from my hand , both me the ball will be moving with velocity of 50 m /s in horizontal direction .

As soon as ball is separated from the hand , it acquires addition velocity in upward direction and acceleration in downward direction . This will give relative velocity to the ball with respect to me . So I will see the ball going in upward direction under  gravitational acceleration . It appears as if I am sitting at rest and ball is going in upward direction under deceleration . My motion at 50 m/s will have no effect on the motion of ball in upward direction , according to first law of Newton . It is so because ball too will be moving in forward direction with the same speed which will not be visible to me because I too am moving with the same speed.

If I  am  sitting at rest at home and I threw a ball straight up into the air , I will have the same experience of seeing ball going in similar way as described above.

8 0
3 years ago
.A coin rolls off the edge of a table. The coin
geniusboy [140]

Answer:

Apply the following formulae horizontally And get A value for time

Remember horizontal acceleration is zero

s  = ut +  \frac{1}{2}a {t}^{2}   \\ 0.8 = 1.7 \times t \\  \frac{0.8}{1.7}  = t \\ t = 0.47s

and then to find the height apply the same above equation vertically...remember vertical initial velocity is zero

s = ut +  \frac{1}{2} a {t}^{2}  \\ s =  \frac{1}{2}  \times 10 \times (0.47) ^{2}  \\ s = 1.1045m

5 0
3 years ago
If Scobie could drive a Jetson's flying car at a constant speed of 160.0 km/hr across oceans and space, approximately how long w
DochEvi [55]

Scobie will take 10 days to drive around Earth's equator.

To calculate the time that takes Scobie to drive around Earth's equator we need to find the distance, which is given by the equation of a circumference:

d = 2\pi r

<em>Where:</em>

r: is the Earth's radius = 6371 km

Then, the distance is:

d = 2\pi r = 2\pi*6371 km = 40030.2 km

Now, if we divide the above distance by the speed of the car we can find the time:

t = \frac{d}{v} = \frac{40030.2 km}{160.0 km/h} = 250.2 h*\frac{1 d}{24 h} = 10 d

Therefore, Scobie will take 10 days to drive around Earth's equator.

     

To learn more about distance and time here: brainly.com/question/14236800?referrer=searchResults

I hope it helps you!

6 0
3 years ago
A solenoid with 35 turns per centimeter carries a current I. An electron moves within the solenoid in a circle that has a radius
castortr0y [4]

Answer:

The current of the solenoid is 0.0129 A.

Explanation:

The movement of the electron within the solenoid in a circle is produced by equaling the magnetic force and the centripetal force, as follows:

F_{B} = F_{c}

e*v \mu_{0}*n*I = \frac{m*v^{2}}{r}

I = \frac{m*v}{e* \mu_{0}*n*r}

Where:

I: is the current

m: is the electron's mass = 9.1x10⁺³¹ kg

v: is the electron's speed = 3.0x10⁵ m/s

μ₀: is the permeability magnetic = 4πx10⁻⁷ T.m/A

n: is the number of turns per unit length = 35/cm

r: is the radius of the circle = 3.0 cm

e: is the electron's charge = 1.6x10⁻¹⁹ C  

I = \frac{m*v}{e*\mu_{0}*n*r} = \frac{9.1 \cdot 10^{-31} kg*3.0 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*4\pi \cdot 10^{-7} T.m/A*3500/m*0.03 m} = 0.0129 A  

Therefore, the current of the solenoid is 0.0129 A.

I hope it helps you!

3 0
4 years ago
If the coefficient of kinetic friction between tires and dry pavement is 0.84, what is the shortest distance in which you can st
liberstina [14]

Answer:

The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

Explanation:

Given;

coefficient of kinetic friction, μ = 0.84

speed of the automobile, u = 29.0 m/s

To determine the  the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;

v² = u² + 2ax

where;

v is the final velocity

u is the initial velocity

a is the acceleration

x is the shortest distance

First we determine a;

From Newton's second law of motion

∑F = ma

F is the kinetic friction that opposes the motion of the car

-Fk = ma

but, -Fk = -μN

-μN = ma

-μmg = ma

-μg = a

- 0.8 x 9.8 = a

-7.84 m/s² = a

Now, substitute in the value of a in the equation above

v² = u² + 2ax

when the automobile stops, the final velocity, v = 0

0 = 29² + 2(-7.84)x

0 = 841 - 15.68x

15.68x = 841

x = 841 / 15.68

x = 53.64 m

Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

4 0
3 years ago
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