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3 years ago

14

a stone is thrown vertically upwards with an initial velocity of 20 metre per second find the maximum height it reaches and the

time taken by to reach the height( g =10m/s)

1 answer:

ziro4ka [17]3 years ago

4 0

Answer:

y = 20 [m]

Explanation:

To solve this type of problem we must use the following equation of kinematics.

$v_{f}^{2} =v_{o}^{2}-2*g*y$

where:

Vf = final velocity = 0

Vo = initial velocity = 20 [m/s]

g = gravity acceleration = 10 [m/s²]

y = elevation [m]

Now the key to solving this problem is to know that when the stone reaches the maximum height its speed is zero. Therefore the final velocity term is zero.

$0 = (20)^{2}-2*10*y\\20*y=400\\y=20 [m]$

Another important fact is that the sign of gravitational acceleration was taken as negative, since it acts in the opposite direction to the movement of the stone.

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Newtons first law 1 to 5. <br>What is each of the net force for all of the 5 questions?

erica [24]

Answer:

1. 65 N.

2. 160 N.

3. 0 N.

4. 210 N.

5. 90 N.

Explanation:

1. Determination of the net force.

Force applied to the right (Fᵣ) = 80 N

Force applied to the left (Fₗ) = 145 N

Net force (Fₙ) =?

Fₙ = Fₗ – Fᵣ

Fₙ = 145 – 80

Fₙ = 65 N

Thus, the net force is 65 N

2. Determination of the net force.

Force 1 applied to the left (F₁) = 35 N

Force 2 applied to the left (F₂) = 125 N

Net force (Fₙ) =?

Fₙ = F₁ + F₂

Fₙ = 35 + 125

Fₙ = 160 N

Thus, the net force is 160 N.

3. Determination of the net force.

Force applied to the right (Fᵣ) = 75 N

Force applied to the left (Fₗ) = 75 N

Net force (Fₙ) =?

Fₙ = Fₗ – Fᵣ

Fₙ = 75 – 75

Fₙ = 0

Thus, the net force is 0 N

4. Determination of the net force.

Force 1 applied to the right (F₁) = 150 N

Force 2 applied to the right (F₂) = 60 N

Net force (Fₙ) =?

Fₙ = F₁ + F₂

Fₙ = 150 + 60

Fₙ = 210 N

Thus, the net force is 210 N.

5. Determination of the net force.

Force applied to the right (Fᵣ) = 115 N

Force applied to the left (Fₗ) = 25 N

Net force (Fₙ) =?

Fₙ = Fᵣ – Fₗ

Fₙ = 115 – 25

Fₙ = 90 N

Thus, the net force is 90 N

8 0

3 years ago

Suppose that you have a 680 Ω, a 720 Ω and a 1.20 kΩ resistor. (a) What is the maximum resistance you can obtain by combining th

Delvig [45]

Explanation:

As the given data is as follows.

$R_{1} = 680 \ohm$ ohm $\ohm$ , $R_{2} = 720 \ohm$ ohm,

$R_{3} = 1.2 k\ohm$ = 1200 $\ohm$ (as 1 k ohm = 1000 m)

(a) We will calculate the maximum resistance by combining the given resistances as follows.

Max. Resistance = $R_{1} + R_{2} + R_{3}$

= $(680 + 720 + 1200) \ohm$ ohm

= 2600 ohm

or, = 2.6 $k\ohm$ ohm

Therefore, the maximum resistance you can obtain by combining these is 2.6 $k\ohm$ ohm.

(b) Now, the minimum resistance is calculated as follows.

Min. Resistance = $\frac{1}{R_{1}} + \frac{1}{R_{2}} + \frac{1}{R_{3}}$

= $\frac{1}{680} + \frac{1}{720} + \frac{1}{1200}$

= $3.683 \times 10^{-3}$ ohm

Hence, we can conclude that minimum resistance you can obtain by combining these is $3.683 \times 10^{-3}$ ohm.

3 0

3 years ago

A 1.30-m string of weight 0.0125 N is tied to the ceiling at its upper end, and the lower end supports a weight W. Neglect the v

atroni [7]

Answer:

1. t = 0.0819s

2. W = 0.25N

3. n = 36

4. y(x , t)= Acos[172x + 2730t]

Explanation:

1) The given equation is

$y(x, t) = Acos(kx -wt)$

The relationship between velocity and propagation constant is

$v = \frac{\omega}{k}=\frac{2730rad/sec}{172rad/m}\\\\$

v = 15.87m/s

Time taken, $t = \frac{\lambda}{v}$

$= \frac{1.3}{15.87}\\\\=0.0819 sec$

t = 0.0819s

2)

The velocity of transverse wave is given by

$v = \sqrt{\frac{T}{\mu}}$

$v = \sqrt{\frac{W}{\frac{m}{\lambda}}}$

mass of string is calculated thus

mg = 0.0125N

$m = \frac{0.0125N}{9.8N/s}$

m = 0.00128kg

$\omega = \frac{v^2m}{\lambda}$

$\omega = \frac{(15.87^2)(0.00128)}{1.30}$

$\omega =$ 0.25N

3)

The propagation constant k is

$k=\frac{2\pi}{\lambda}$

hence

$\lambda = \frac{2\pi}{k}\\\\\lambda = \frac{2 \times 3.142}{172}$

$\lambda =$ 0.036 m

No of wavelengths, n is

$n = \frac{L}{\lambda}\\\\n = \frac{1.30m}{0.036m}\\$

n = 36

4)

The equation of wave travelling down the string is

$y(x, t)=Acos[kx -wt]\\\\becomes\\\\y(x , t)= Acos[(172 rad.m)x + (2730 rad.s)t]$

$without, unit\\\\y(x , t)= Acos[172x + 2730t]$

7 0

3 years ago

A force has both______ and _______. Question 1 options: Speed and impact Magnitude and direction Motion and speed Direction and

Travka [436]

Answer:

magnitude

and direction

Explanation:

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2 years ago