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3 years ago

Question 3.Two forces are acting on a 30-kg object, 200 N up and 300 N

Physics

1 answer:

atroni [7]3 years ago

4 0

Answer:

It is 3 m/s² downward.

Explanation:

The mass of an object, m = 30 kg

Upward force, F = 200 N

Downward force, F' = -300 N

We need to find the magnitude and direction of its acceleration.

Net force = F+F'

= 200+(-300)

= -100 N

We know that,

Force, F = ma, a is acceleration of the object

$a=\dfrac{F}{m}\\\\a=\dfrac{-100}{30}\\\\a=-3.33\ m/s^2\approx -3\ m/s^2$

Hence, the acceleration is -3 m/s² and it is acting in the downward direction.

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The acceleration due to gravity on jupiter is 2.5 times whats on earth. An object of mass 10kg is taken to Jupiter. What is the

Damm [24]

The mass is still 10 kg. But instead of weighing 98N as it does on Earth, it weighs 245N on Jupiter.

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3 years ago

What is the voltage across each resistor?

kogti [31]

Answer:

see solution below

Explanation:

The given resistors are connected in series.

Equivalent resistance in series = 30 + 55 + 15

Equivalent resistance in series Rt = 100 ohms

Since the potential difference in the circuit = 36V

Get the current in the circuit first

I = V/Rt

I = 36/100

I = 0.36A

Get the voltage across 30ohms resistor;

V30 = 0.36 * 30

V30 = 10.8volts

Hence the voltage across the 30ohms resistor is 10.8volts

Get the voltage across 55ohms resistor;

V55 = 0.36 * 55

V55 = 19.8volts

Hence the voltage across the 55ohms resistor is 19.8volts

Get the voltage across 15ohms resistor;

V15 = 0.36 * 15

V15 = 5.4volts

Hence the voltage across the 15ohms resistor is 5.4volts

4 0

3 years ago

(a) Calculate the force (in N) needed to bring a 1100 kg car to rest from a speed of 85.0 km/h in a distance of 125 m (a fairly

nasty-shy [4]

(a) -2451 N

We can start by calculating the acceleration of the car. We have:

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 125 m is the stopping distance

So we can use the following equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car, a:

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(125 m)}=-2.23 m/s^2$

Now we can use Newton's second Law:

F = ma

where m = 1100 kg to find the force exerted on the car in order to stop it; we find:

$F=(1100 kg)(-2.23 m/s^2)=-2451 N$

and the negative sign means the force is in the opposite direction to the motion of the car.

(b) $-1.53\cdot 10^5 N$

We can use again the equation

$v^2 - u^2 = 2ad$

To find the acceleration of the car. This time we have

$u=85.0 km/h = 23.6 m/s$ is the initial velocity

v = 0 is the final velocity of the car

d = 2.0 m is the stopping distance

Substituting and solving for a,

$a=\frac{v^2-u^2}{2d}=\frac{0-(23.6 m/s)^2}{2(2 m)}=-139.2 m/s^2$

So now we can find the force exerted on the car by using again Newton's second law:

$F=ma=(1100 kg)(-139.2 m/s^2)=-1.53\cdot 10^5 N$

As we can see, the force is much stronger than the force exerted in part a).

8 0

3 years ago

If 0.035pC of charge is transferred via the movement of Al3+ ions, how's many of these must be transferred in total? Please add

mr Goodwill [35]

Each $Al^+^3$ ion contains three extra protons. Hence, the extra charge on each $Al^+^3$ = $3 \times 1.6 \times 10^-^1^9$ C

Total charge = 0.035 pC

Total charge (Q) = $0.035 \times 10^-^1^2$ C

Let the number of $Al^+^3$ ions be n.

According to question:

$n \times 3 \times 1.6 \times 10^-^1^9 =0.035 \times 10^-^1^2$

$n = \frac{0.035 \times 10^-^1^2}{3 \times 1.6 \times 10^-^1^9}$

$n = 7.29167 \times 10^4$

n = 72917

Hence, the total number of ions needed to be transferred is 72917

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3 years ago

A particle is moving along a projectile path where the initial height is 160 feet with an initial speed of 144 feet per second.

Leviafan [203]

Assuming Earth's gravity, the formula for the flight of the particle is:

s(t) = -16t^2 + vt + s = -16t^2 + 144t + 160. 

This has a maximum when t = -b/(2a) = -144/[2(-16)] = -144/(-32) = 9/2. 

Therefore, the maximum height is s(9/2) = -16(9/2)^2 + 144(9/2) + 160 = 484 feet.

7 0

3 years ago

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