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Pepsi [2]
3 years ago
11

Three collinear forces,F1=45N west,F2=63N east and an unknown force F3 are applied to an object.The resultant force of the three

forces is 12N west.
Show by means of a tail to head method of vector addition how the magnitude of F3 could be found?​
Physics
1 answer:
Greeley [361]3 years ago
6 0

Take east to be the positive direction. Then the resultant force from adding <em>F</em>₁ and <em>F</em>₂ is

<em>F</em>₁ + <em>F</em>₂ = (-45 N) + 63 N = 18 N

which is positive, so it's directed east.

To this we add a third force <em>F</em>₃ such that the resultant is 12 N pointing west, making it negative, so that

18 N + <em>F</em>₃ = -12 N

<em>F</em>₃ = -30 N

So <em>F</em>₃ has a magnitude of 30 N and points west.

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A family car has a mass of 1400 kg. In an accident it hits a wall and goes from a speed of 27 m/s to a standstill in 1.5 seconds
horrorfan [7]

Answer:

The force has been reduced by 8018 N

Explanation:

The impulse exerted on the car during the crash is equal to the product of the force exerted and the duration of the collision, and it is also equal to the change in momentum of the car. So we can write:

F\Delta t = m\Delta v

where:

F is the force exerted on the car

\Delta t is the duration of the collision

m = 1400 kg is the mass of the car

\Delta  v=-27 m/s is the change in velocity of the car

We can re-write the equation as

F=\frac{m\Delta v}{\Delta t}

In the 1st collision, the time is 1.5 seconds, so the force is

F_1=\frac{(1400)(-27)}{1.5}=-25,200 N

In the 2nd collision, the time is increased to 2.2 seconds, so the force is

F_2=\frac{(1400)(-27)}{2.2}=-17,182 N

Therefore, the force has been reduced by:

F_2-F_1=-17,182-(-25,200)=8018 N

4 0
4 years ago
Read 2 more answers
It is easier to drive In sharp nails in a piece of wood than the blunt ones why?
murzikaleks [220]

The sharp nail has a less surface area in comparison to a blunt nail and pressure is inversely proportional to area so it is easier to Hamer a sharp  nail into a wood rather than having a blunt nail in wood


5 0
3 years ago
A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy tra
kkurt [141]

Answer:

V_{ft}= 317 cm/s

ΔK = 2.45 J

Explanation:

a) Using the law of the conservation of the linear momentum:

P_i = P_f

Where:

P_i=M_cV_{ic} + M_tV_{it}

P_f = M_cV_{fc} + M_tV_{ft}

Now:

M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}

Where M_c is the mass of the car, V_{ic} is the initial velocity of the car, M_t is the mass of train, V_{fc} is the final velocity of the car and V_{ft} is the final velocity of the train.

Replacing data:

(1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}

Solving for V_{ft}:

V_{ft}= 3.17 m/s

Changed to cm/s, we get:

V_{ft}= 3.17*100 = 317 cm/s

b) The kinetic energy K is calculated as:

K = \frac{1}{2}MV^2

where M is the mass and V is the velocity.

So, the initial K is:

K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2

K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2

K_i = 22.06 J

And the final K is:

K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2

K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2

K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2

K_f = 19.61 J

Finally, the change in the total kinetic energy is:

ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J

4 0
4 years ago
3. A motorbike travels 45 miles in 15 minutes, what is its speed?
antiseptic1488 [7]
Hes going 180 mph
oof...
3 0
3 years ago
How to find acceleration in non uniform motion. Please give two ways
Olenka [21]

<u>The two ways to  find acceleration in non uniform motion are as follows:</u>

  • By Graphs
  • By Calculus

<u>Explanation:</u>

Non-uniform acceleration comprises the most common description of motion. Acceleration refers to the rate of changes of velocity per unit time. Basically, it implies that acceleration changes during motion. This variety can be communicated either as far as position (x) or time (t).

Accordingly, non-uniform acceleration motion can be carried out in 2 ways:

  • By Calculus
  • By graphs

Calculus analysis is general and accurate, but limited to the availability of speed and acceleration expressions. It is not always possible to get the expression of motion attributes in the form "x" or "t". On the other hand, the graphic method is not accurate enough, but it can be used accurately if the graphic has the correct shapes.

The use of calculations involves differentiation and integration. Integration enables evaluation of the expression of acceleration of speed and expression of movement at a distance. Similarly, differentiation allows us to evaluate expression of speed position and expression speed to acceleration.

6 0
3 years ago
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