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2 years ago

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These three bulbs are powered by the battery. What will happen if the middle light burns out? A. The two other bulbs will go out

. B. The two other bulbs will stay on. C. Just the bulb closest to the battery will stay on. D. Just the bulb farthest from the battery will stay on.

1 answer:

Anna11 [10]2 years ago

4 0

Answer:

C

Explanation:

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A rocket is launched from rest and moves in a straight line at 30.0 degrees above the horizontal with an acceleration of 35.0 m/

klemol [59]

Answer:

t = 123.59s

Explanation:

For the launch pad section:

Vf = Vo + a*t where Vo=0.

Vf = 35*25 = 875m/s

The distance traveled during the launch:

$d = Vo*t+\frac{a*t^2}{2} = 0+\frac{35*25^2}{2} = 10937.5m$

Now the projectile motion, we know that its initial speed is the speed calculated previously and the initial height is the y-component of the previously calculated distance.

$\Delta Y = Vo*sin(30)*t - \frac{g*t^2}{2}$

$-d*sin(30) = Vo*sin(30)*t - \frac{g*t^2}{2}$ where d= 10937.5m; Vo=875m/s.

Solving for t:

t1 = -11.093s t2 = 98.59s

So, the total time of flight will be:

$t_{total} = t_{launch}+t_{projectile}=25+98.59 = 123.59s$

6 0

2 years ago

A boy takes hold of a rope to pull a wagon (m = 50 kg) on a surface with a static coefficient of friction μS = 0.25. Calculate t

vivado [14]

Answer:

<em>The force that would be applied on the rope just to start moving the wagon is 122 N</em>

Explanation:

Frictional force opposes motion between two surfaces in contact. It is the force that must be applied before a body starts to move. Static friction opposes the motion of two bodies that are in contact but are not moving. The magnitude of static friction to overcome for the body to move can be calculated using equation 1.

F = μ x mg .............................. 1

where F is the frictional force;

μ is the coefficient of friction ( μs, in this case, static friction);

m is mass of the object and;

g is the acceleration due to gravity( a constant equal to 9.81 m/ $s^{2}$ )

from the equation we are provide with;

μs = 0.25

m = 50 kg

g = 9.81 m/ $s^{2}$

F =?

Using equation 1

F = 0.25 x 50 kg x 9.81 m/ $s^{2}$

F = 122.63 N

<em>Therefore a force of 122 N must be applied to the rope just to start the wagon.</em>

6 0

2 years ago

Read 2 more answers

If the satellite has a mass of 3900 kg , a radius of 4.3 m , and the rockets each add a mass of 210 kg , what is the required st

ehidna [41]

Answer:

The required steady force of each rocket is 28.79 N

Explanation:

mass of the satellite, M=3900 kg

radius, r=4.3 m

mass of rocket, m=210 kg

time, t=5.4 min

Moment of Inertia:

I = 1/2 (Mr^2) + 4mr^2

I = 1/2 ( 3900* (4.3)^2) + 4 (210)*(4.3)^2

I = 51587.1 kg m^2

the angular acceleration is:

a= w/t

here w= 2*π*30

so,

a= 2*π*30 / 5.4* 3600

a=0.0096 rad/ s^2

the Torque becomes:

T=I*a = 4r*F

( 51587.1 )*(0.0096) = 4*4.3* F

F= 28.79 N

the required steady force of each rocket is 28.79 N

learn more about steady force here:

<u>brainly.com/question/13841147</u>

<u />

#SPJ4

7 0

1 year ago

Suppose the maximum power delivered by a car's engine results in a force of 16000 N on the car by the road. In the absence of an

joja [24]

Answer:

Approximately $9.7\; \rm m \cdot s^{-2}$ .

Explanation:

Assuming that there is no other force on this vehicle, the $16000\; \rm N$ force from the road would be the only force on this vehicle. The net force would then be equal to this $16000\; \rm N\!$ force. The size of the net force would be $16000\; \rm N\!\!$ .

Let $m$ denote the mass of this vehicle and let $\Sigma F$ denote the net force on this vehicle.

By Newton's Second Law of motion, the acceleration of this vehicle would be proportional to the net force on this vehicle. In other words, the acceleration of this vehicle, $a$ , would be:

$\begin{aligned}a &= \frac{\Sigma F}{m}\end{aligned}$ .

For this vehicle, $\Sigma F = 16000\; \rm N$ whereas $m = 1650\; \rm kg$ . The acceleration of this vehicle would be:

$\begin{aligned}a &= \frac{16000\; \rm N}{1650\; \rm kg} \\ &= \frac{16000\; \rm kg \cdot m\cdot s^{-2}}{1650\; \rm kg}\\ &\approx 9.7 \; \rm m \cdot s^{-2}\end{aligned}$ .

8 0

2 years ago

All fires require oxygen () There is no oxygen in that room (T) .

Marina CMI [18]

The sentences are invalid and unsound.

<h3><u>Explanation</u>:</h3>

The fire is defined as the vigorous oxidation of a substance. Now oxidation can occur in presence of any oxidising agent. Like magnesium in presence of nitrogen in high temperature with a dazzling brownish flame to produce magnesium nitride. So fire can be produced in absence of oxygen.

Oxygen is present everywhere in world. So production of a whole room without oxygen is very tough to produce and costly process. So its very unsound.

4 0

2 years ago