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3 years ago

13

Gravitational pull is determined by ______?

1 answer:

IgorLugansk [536]3 years ago

6 0

The answer is D. mass

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Particles have an orderly arrangement in this type of solid. Name this type of solid

AnnyKZ [126]

Answer:

Solid crystal

Explanation:

6 0

4 years ago

Alex goes cruising on his dirt bike. He rides 700 m north, 300 m east, 400 m north, 600 m west, 1200 m south, 300 m east, and fi

velikii [3]

ANSWER

$0\operatorname{km}$

EXPLANATION

First, let us make a sketch of the question:

From the diagram:

=> black line represents North

=> green line represents East

=> blue line represents West

=> red line represents South

From the diagram, we see that at the end of his journey, he returns to his start point.

Since displacement is the measure of the change in the position of an object and the boy's position did not change after the journey, his displacement is:

$0\operatorname{km}$

7 0

1 year ago

The Great Sandini is a 60 kg circus performer who is shot from a cannon (actually a spring gun). You don't find many men of his

Alex17521 [72]

Answer:

22m/s

Explanation:

Mass, m=60 kg

Force constant, k=1300N/m

Restoring force, Fx=6500 N

Average friction force, f=50 N

Length of barrel, l=5m

y=2.5 m

Initial velocity, u=0

$F_x=kx$

Substitute the values

$6500=1300x$

$x=\frac{6500}{1300}=5$ m

Work done due to friction force

$W_f=fscos\theta$

We have $\theta=180^{\circ}$

Substitute the values

$W_f=50\times 5cos180^{\circ}$

$W_f=-250J$

Initial kinetic energy, Ki=0

Initial gravitational energy, $U_{grav,1}=0$ \

Initial elastic potential energy

$U_{el,1}=\frac{1}{2}kx^2=\frac{1}{2}(1300)(5^2)$

$U_{el,1}=16250J$

Final elastic energy, $U_{el,2}=0$

Final kinetic energy, $K_f=\frac{1}{2}(60)v^2=30v^2$

Final gravitational energy, $U_{grav,2}=mgh=60\times 9.8\times 2.5$

Final gravitational energy, $U_{grav,2}=1470J$

Using work-energy theorem

$K_i+U_{grav,1}+U_{el,1}+W_f=K_f+U_{grav,2}+U_{el,2}$

Substitute the values

$0+0+16250-250=30v^2+1470+0$

$16000-1470=30v^2$

$14530=30v^2$

$v^2=\frac{14530}{30}$

$v=\sqrt{\frac{14530}{30}}$

$v=22m/s$

5 0

3 years ago

Parallel conducting tracks, separated by 2.20 cm, run north and south. There is a uniform magnetic field of 1.20 T pointing upwa

mars1129 [50]

Answer:

The magnitude of the magnetic force on the rod is 0.037 N.

Explanation:

The magnetic force is given by:

$F = qvBsin(\theta)$

Since the charge (q) is:

$q = I*t$

Where<em> I</em> is the current = 1.40 A, and <em>t</em> the time

And the speed (v):

$v = \frac{L}{t}$

Where <em>L </em>is the tracks separation = 2.20 cm = 0.022 m

Hence, the magnetic force is:

$F = ILBsin(\theta)$

Where <em>B </em>is the magnetic field = 1.20 T and <em>θ</em> is the angle between the tracks and the magnetic field = 90°

$F = ILBsin(\theta) = 1.40*0.022*1.20*sin(90) = 0.037 N$

Therefore, the magnitude of the magnetic force on the rod is 0.037 N.

I hope it helps you!

5 0

3 years ago

Cars cross a certain point on the highway in accordance with a Poisson process with rate = 3 per minute. If Al runs across the h

Svetradugi [14.3K]

Answer:

The probability is 0.2212

Solution:

As per the question:

Poisson rate, $\lambda = 3/min = \frac{3}{60}\ s = 0.05\ s$

Now,

Let

X: time of waiting for the next vehicle on the highway

Now,

To find the probability of the next vehicle to arrive after 10 s

The probability distribution function is given by:

$f(x) = \lambda e^{- \lambda x}$

Now,

P(X < x) = $1 - e^{- \lambda x}$

$P(X \leq x) = 1 - e^{- 0.05 x}$

For X> 0,

P(X > x) = $e^{- \lambda x}$

P(X < 5) = $1 - e^{- 5\lambda} = 1 - e^{- 0.25} = 1 - 0.7788 = 0.2212$

4 0

3 years ago