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aleksandrvk [35]
3 years ago
8

Help asap plz !!! Use the circuit diagram to decide if the lightbulb will

Physics
1 answer:
nata0808 [166]3 years ago
4 0

Answer:

I wish I knew the answer to that question

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A car is pushed with a force of 450 N for 19.4 seconds. What impulse was applied to the car?​
harina [27]

Answer:

impulse = 8820 kg·\frac{m}{s} or 8820 N·s

Explanation:

Impulse J is equal to the average force F_{av} multiplied by the elapsed time Δt or in equation form, J = F_{av}Δt

As long as your force of 450 N is constant then that value is your average force F_{av} and your elapsed time is 19.4 seconds.

Multiply these values.

You will get an impulse of 8820 kg·\frac{m}{s} or 8820 N·s.

6 0
2 years ago
Toon Train is traveling at the speed of 10 m/s at the top of a hill. Five seconds later it reaches the bottom of the hill and is
Naddika [18.5K]

Answer:

the rate of acceleration of the train is 4 m/s²

Explanation:

Given;

initial velocity of the train, u = 10 m/s

change in time of motion, dt = 5 s

final velocity of the train, v = 30 m/s

The rate of acceleration of the train is calculated as;

a = \frac{dv}{dt} = \frac{v-u}{dt} = \frac{30-10}{5} = \frac{20}{5} = 4 \ m/s^2

Therefore, the rate of acceleration of the train is 4 m/s²

5 0
3 years ago
Lonic Naming Practice<br> Write the name of the following chemical compound:<br> Na3Cl1
GaryK [48]

Answer:

sodium chloride. .....

6 0
3 years ago
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Zero, a hypothetical planet, has a mass of 5.3 x 1023 kg, a radius of 3.3 x 106 m, and no atmosphere. A 10 kg space probe is to
Andrej [43]

(a) 3.1\cdot 10^7 J

The total mechanical energy of the space probe must be constant, so we can write:

E_i = E_f\\K_i + U_i = K_f + U_f (1)

where

K_i is the kinetic energy at the surface, when the probe is launched

U_i is the gravitational potential energy at the surface

K_f is the final kinetic energy of the probe

U_i is the final gravitational potential energy

Here we have

K_i = 5.0 \cdot 10^7 J

at the surface, R=3.3\cdot 10^6 m (radius of the planet), M=5.3\cdot 10^{23}kg (mass of the planet) and m=10 kg (mass of the probe), so the initial gravitational potential energy is

U_i=-G\frac{mM}{R}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{3.3\cdot 10^6 m}=-1.07\cdot 10^8 J

At the final point, the distance of the probe from the centre of Zero is

r=4.0\cdot 10^6 m

so the final potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{4.0\cdot 10^6 m}=-8.8\cdot 10^7 J

So now we can use eq.(1) to find the final kinetic energy:

K_f = K_i + U_i - U_f = 5.0\cdot 10^7 J+(-1.07\cdot 10^8 J)-(-8.8\cdot 10^7 J)=3.1\cdot 10^7 J

(b) 6.3\cdot 10^7 J

The probe reaches a maximum distance of

r=8.0\cdot 10^6 m

which means that at that point, the kinetic energy is zero: (the probe speed has become zero):

K_f = 0

At that point, the gravitational potential energy is

U_f=-G\frac{mM}{r}=-(6.67\cdot 10^{-11})\frac{(10 kg)(5.3\cdot 10^{23}kg)}{8.0\cdot 10^6 m}=-4.4\cdot 10^7 J

So now we can use eq.(1) to find the initial kinetic energy:

K_i = K_f + U_f - U_i = 0+(-4.4\cdot 10^7 J)-(-1.07\cdot 10^8 J)=6.3\cdot 10^7 J

3 0
3 years ago
Compounds that contain only carbon and hydrogen are _____.
REY [17]

Answer:

are known as hydrocarbons. the saturated hydrocarbons are known as alkanes.

5 0
3 years ago
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