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3 years ago

6

A particle leaves the origin with an initial velocity v = (3.00 i hat) m/s and a constant acceleration a = (-3.00 i hat - 1.400

j hat) m/s2
(a) What is the velocity of the particle when it reaches its maximum x coordinate?

m/s i hat + m/s j hat + m/s k hat

(b) What is the position vector of the particle at this time?

m i hat + m j hat + m k hat

1 answer:

Brilliant_brown [7]3 years ago

8 0

Answer:

a) —0.5 j m/s

b) 4.5 i + 2.25 j m

Explanation:

<u>Givens:</u>

v_0 =3.00 i m/s

a= (-3 i — 1.400 j ) m/s^2

The maximum x coordinate is reached when dx/dt = 0 or v_x = 0 ,thus :

<em>v_x = v_0 + at = 0 </em>

(3.00 i m/s) + (-3 i m/s^2)t=0

t = (3 m/s)/-3 i m/s^2

t = -1 s

Therefore the particle reaches the maximum x-coordinate at time t = 1 s.

Part a The velocity-of course- is all in the y-direction,therefore:

v_y =v_0+ at

We have that v_0 = 0 in the y-direction.

v_y = (-0.5 j m/s^2)(1 s)

= —0.5 j m/s

Part b: While the position of the particle at t = 1 s is given by:

r=r_0+v_0*t+1/2*a*t^2

Where r_0 = 0 since the particle started from the origin.

Its position at t = 1 s is then given by :

r =(3.00 i m/s)(1 s)+1/2(-3 i — 1.400 j )(1 s)^2

=4.5 i + 2.25 j m

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Answer:9.75 m/s

Explanation:

Given

Length of ladder $(L)=5 m$

Foot the ladder is moving away with speed of $\frac{\mathrm{d} x}{\mathrm{d} t}=13 m/s$

From diagram

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$y^2=25-9=16$

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Now differentiating equation 1 w.r.t time

$2x\frac{\mathrm{d} x}{\mathrm{d} t}+2y\frac{\mathrm{d} y}{\mathrm{d} t}=0$

$x\frac{\mathrm{d} x}{\mathrm{d} t}=-y\frac{\mathrm{d} y}{\mathrm{d} t}$

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3 years ago

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Dmitriy789 [7]

Answer:

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3 years ago

The resistance and the magnitude of the current depend on the path that the current takes. The drawing shows three situations in

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Answer: (a). Resistance = 0.4286ohms and Current (I) = 7A

(b). Resistance (R) = 0.027 ohms and Current (I) = 111.1A

(c). Resistance (R) = 0.1071 ohms and Current (I) = 28A

Explanation:

From the question, given that;

ρ = 1.5*10-2ῼ.m

Lo = 7cm = 0.07m

V = 3V

From the formula R = ρL/A, where A is the area of cross section, L is the length of material and ρ is the resistivity.

(A)

L = 4Lo and A = 2Lo*Lo

R = ρL/A

R = ρ4Lo/(2Lo*Lo)

R = 2ρ /Lo = 2*1.5*10-2/0.07

R = 0.4286 ῼ

From this the current becomes;

I = V/R = 3/ 0.4286 = 6.99 = 7A

(B)

L = Lo and A = 4Lo * 2Lo

R = ρL/A

R = ρLo/ (4Lo*2Lo) after eliminating Lo from both sides we get,

R = ρ/8Lo = 1.5*10-2 / 8*0.07

R = 0.027

Current (I) = V/R = 3/0.027 = 111.1A

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L = 2Lo and A = Lo * 4Lo

R = ρL/A

R = ρ2Lo/ (Lo*4Lo) eliminating Lo from both sides we get,

R = ρ/2Lo = 1.5*10-2 / 2*0.07 = 0.1071

The current becomes;

I = V/R = 3/0.1071 = 28A

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4 years ago

Read 2 more answers

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TiliK225 [7]

Answer:

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(c)By $v^2 = u^2 - 2gh$ => H = 46045.92 m.

Explanation:

a) By $v^2 = u^2 + 2as$

$(950)^2 = 0 + 2 \times a \times 0.75\\a = 601666.67 m/s^2\\a/g = 688858.70/9.8 = 70291.70$

(b)By $v = u + at$

$950 = 0 + 601666.67 \times t\\t = 1.58 x 10^-3 sec = 1.58 ms$

(c)By $v^2 = u^2 - 2gh$

$0 = (950)^2 - 2 \times9.8 \times H\\H = 46045.92 m$

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3 years ago

A ball filled with an unknown material starts from rest at the top of a 2 m high incline that makes a 28o with respect to the ho

Lady_Fox [76]

Answer:

<u>Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!</u>

The ball rotates 6.78 revolutions.

Explanation:

<u>Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!</u>

At the bottom the ball has the following angular speed:

$\omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s$

Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:

$sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m$

To find the revolutions we need the time, which can be found using the following equation:

$v_{f} = v_{0} + at$

$t = \frac{v_{f} - v_{0}}{a}$ (1)

So first, we need to find the acceleration:

$v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L}$ (2)

By entering equation (2) into (1) we have:

$t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}}$

Since it starts from rest (v₀ = 0):

$t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s$

Finally, we can find the revolutions:

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Therefore, the ball rotates 6.78 revolutions.

I hope it helps you!

3 0

3 years ago