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horsena [70]
3 years ago
7

Please Help!

Physics
1 answer:
Mamont248 [21]3 years ago
7 0
In the parallel circut three bulbs are connected  . if the one bulbs is 1 amps three are equal so yhe answer is 3 amps cause 1+1+1 EQUALS 3 AMPS 

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7. A stick of length L and mass M is hanging at rest from its top edge from a ceiling hinged at that point so that it is free to
Lilit [14]

Answer:

The distance from the top of the stick would be 2l/3

Explanation:

Let the impulse 'FΔt' acts as a distance 'x' from the hinge 'H'. Assume no impulsive reaction is generated at 'H'. Let the angular velocity of the rod about 'H' just after the applied impulse be 'W'. Also consider that the center of percussion is the point on a bean attached to a pivot where a perpendicular impact will produce no reactive shock at the pivot.

Applying impulse momentum theorem for linear momentum.

FΔt = m(Wl/2), since velocity of center of mass of rod  = Wl/2

Similarly applying impulse momentum theorem per angular momentum about H

FΔt * x = I * W

Where FΔt * x represents the impulsive torque and I is the moment of inertia

F Δt.x = (ml² . W)/3

Substituting FΔt

M(Wl/2) * x = (ml². W)/3

1/x = 3/2l

x = 2l/3

8 0
3 years ago
magine two carts, one with twice the mass of the other, that are going to have a head-on collision. In order for the two carts t
scoray [572]

Answer:

Twice as fast

Explanation:

Solution:-

- The mass of less massive cart = m

- The mass of Massive cart = 2m

- The velocity of less massive cart = u

- The velocity of massive cart = v

- We will consider the system of two carts to be isolated and there is no external applied force on the system. This conditions validates the conservation of linear momentum to be applied on the isolated system.

- Each cart with its respective velocity are directed at each other. And meet up with head on collision and comes to rest immediately after the collision.

- The conservation of linear momentum states that the momentum of the system before ( P_i ) and after the collision ( P_f ) remains the same.

                             P_i = P_f

- Since the carts comes to a stop after collision then the linear momentum after the collision ( P_f = 0 ). Therefore, we have:

                             P_i = P_f = 0

- The linear momentum of a particle ( cart ) is the product of its mass and velocity as follows:

                             m*u - 2*m*v = 0

Where,

                 ( u ) and ( v ) are opposing velocity vectors in 1-dimension.

- Evaluate the velcoity ( u ) of the less massive cart in terms of the speed ( v ) of more massive cart as follows:

                          m*u = 2*m*v

                              u = 2*v

Answer: The velocity of less massive cart must be twice the speed of more massive cart for the system conditions to hold true i.e ( they both come to a stop after collision ).

8 0
3 years ago
Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t
Readme [11.4K]

Answer: 3.66(10)^{33}kg

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity \omega of the planet P1 with a period T=750years=2.36(10)^{10}s:

\omega=\frac{2\pi}{T}=\frac{V_{1}}{R} (1)

Where:

V_{1}=40.2km/s=40200m/s is the velocity of planet P1

R is the radius of the orbit of planet P1

Finding R:

R=\frac{V_{1}}{2\pi}T (2)

R=\frac{40200m/s}{2\pi}2.36(10)^{10}s (3)

R=1.5132(10)^{14}m (4)

On the other hand, we know the gravitational force F between the star S with mass M and the planet P1 with mass m is:

F=G\frac{Mm}{R^{2}} (5)

Where G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

In addition, the centripetal force F_{c} exerted on the planet is:

F_{c}=\frac{m{V_{1}}^{2}}{R^{2}} (6)

Assuming this system is in equilibrium:

F=F_{c} (7)

Substituting (5) and (6) in (7):

G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}} (8)

Finding M:

M=\frac{V^{2}R}{G} (9)

M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}} (10)

Finally:

M=3.66(10)^{33}kg (11) This is the mass of the star S

4 0
4 years ago
There are four springs stretched by the same mass.
kirill115 [55]

Answer:

That would be Spring C

Explanation:

i took the test :P

6 0
4 years ago
What do all elements in a column in the periodic table have in common?
JulsSmile [24]

Answer:

1, their atoms have the same number of valence electron. because valence electron determine the group of elements.

6 0
3 years ago
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