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Lera25 [3.4K]
1 year ago
12

The chart shows data for four moving objects. A 4 column table with 4 rows. The first column is labeled Object with entries, W,

X, Y, Z. The second column is labeled Initial Velocity in meters per second with entries, 11, 10, 12, 20. The third column is labeled Final Velocity in meters per second with entries, 29, 34, 40, 28. The fourth column is labeled Change in Time in seconds with entries, 6, 12, 7, 8. Which object has the greatest acceleration? W X Y Z
Physics
1 answer:
KatRina [158]1 year ago
3 0

Answer:

y

Explanation:

I took the test

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The voltage across the terminals of a 9.0 v battery is 8.5 v when the battery is connected to a 60 ω load. part a what is the ba
snow_lady [41]
Refer to the diagram shown below.

i = the current in the circuit., A
R₁ = the internal resistance of the battery, Ω
R₂ = the resistance of the 60 W load, Ω

Because the resistance across the battery is 8.5 V instead of 9.0 V, therefore
(R₁ )(i A) = 9 - 8.5 = (0.5 V)
R₁*i = 0.5         (10

Also,
R₂*i = 9.5         (2)

Because the power dissipated by R₂ is 60 W, therefore
i²R₂ = 60
From (2), obtain
i*9.5 = 60
i = 6.3158 A

From (1), obtain
6.3158*R₁ = 0.5
R₁ = 0.5/6.3158 = 0.0792 Ω = 0.08 Ω (nearest hundredth)

Answer: 0.08 Ω

3 0
1 year ago
Read 2 more answers
A satellite revolves around a planet at an altitude equal to the radius of the planet. the force of gravitational interaction be
USPshnik [31]
<span>f2 = f0/4 The gravity from the planet can be modeled as a point source at the center of the planet with all of the planet's mass concentrated at that point. So the initial condition for f0 has the satellite at a distance of 2r, where r equals the planet's radius. The expression for the force of gravity is F = G*m1*m2/r^2 where F = Force G = Gravitational constant m1,m2 = masses involved r = distance between center of masses. Now for f2, the satellite has an altitude of 3r and when you add in the planet's radius, the distance from the center of the planet is now 4r. When you compare that to the original distance of 2r, that will show you that the satellite is now twice as far from the center of the planet as it was when it started. So let's compare the gravitational attraction, before and after. f0 = G*m1*m2/r^2 f2 = G*m1*m2/(2r)^2 f2/f0 = (G*m1*m2/(2r)^2) / (G*m1*m2/r^2) The Gm m1, and m2 terms cancel, so f2/f0 = (1/(2r)^2) / (1/r^2) f2/f0 = (1/4r^2) / (1/r^2) And the r^2 terms cancel, so f2/f0 = (1/4) / (1/1) f2/f0 = (1/4) / 1 f2/f0 = 1/4 f2 = f0*1/4 f2 = f0/4 So the gravitational force on the satellite after tripling it's altitude is one fourth the original force.</span>
6 0
1 year ago
A 1.0-m-long copper wire of diameter 0.10 cm carries a current of 50.0 A to the east. Suppose we apply to this wire a magnetic f
Setler79 [48]

Answer:

The classification of that same issue in question is characterized below.

Explanation:

The given values are:

Current, I = 50.0 A

Diameter, d = 0.10 cm

(a)...

As we know,

⇒  Magnetic force = Copper wire's weight

So,

⇒   B\times I\times L=M\times g

On putting the estimated values, we get

⇒  B\times 50\times 1=7.037\times 10^{-3}\times 9.81

⇒  50B=69.03297\times 10^{-3}

⇒  B=1.38\times 10^{-3} \ T

(b)...

As we know,

⇒  m=\delta\times L\times \frac{\pi \ d^2}{4}

⇒      =8960\times 1\times \frac{\pi \ (0.001)^2}{4}

⇒      =2240\times \pi \ 0.000001

⇒      =7.037\times 10^{-3} \ kg

7 0
2 years ago
A model train engine was moving at a constant speed on a straight horizontal track. As the engine moved​ along, a marble was fir
bagirrra123 [75]

Answer:

The marble was moving in a projectile and the speed of the engine was 2.716 m/s

Explanation:

The vertical component of the marble's flight path relative to the train

is given by the equation y(t) = v*t - (4.9)*t^2,

where, v is the initial upward velocity of the marble relative to the train.  

So with y(1) = v - 4.9 = 0 we have  

v = 4.9 m/s.

The marble will reach maximum height after 0.5 seconds, at which the

height will be y(0.5) = (4.9)*(0.5) - (4.9)*(0.5)^2 = (4.9)*(0.25) = 1.225 m.

Now,  the marble has a vertical velocity component of 4.9 m/s and a horizontal velocity component

of V m/s such that tan(61) = 4.9 / V

V = 4.9 / tan(61) = 2.716 m/s

This horizontal velocity component of the marble is the same as the

speed of the train i.e. 2.716 m/s.

3 0
2 years ago
A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the
lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

l^2 = x^2 + y^2-----(1)

Differentiating with respect to t ( time ),

0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}  ( l = 26 feet = constant )

\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}

\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}

We have,

y = 10, \frac{dx}{dt}= -3\text{ feet per min}

\frac{dy}{dt}=\frac{3x}{10}-----(X)

(i) From equation (1),

26^2 = x^2 + 10^2

676=x^2 + 100

576 = x^2

\implies x = 24\text{ feet}

From equation (X),

\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}

(ii) From equation (X),

\frac{dy}{dt}\propto x

Thus, for different value of x the value of \frac{dy}{dt} would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

\frac{dy}{dt}=0\text{ feet per min}

3 0
2 years ago
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