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lana [24]
3 years ago
6

A record of travel along a straight path is as follows: 1. Start from rest with constant acceleration of 2.65 m/s2 for 17.0 s. 2

. Maintain a constant velocity for the next 1.60 min. 3. Apply a constant negative acceleration of −9.39 m/s2 for 4.80 s.
(a) What was the total displacement for the trip?

b) What were the average speeds for legs 1, 2, and 3 of the trip, as well as for the complete trip
Physics
1 answer:
nlexa [21]3 years ago
8 0
Hello

Let's solve the problem in the three different steps

1) Uniformly accelerated motion, with acceleration a_1 = 2.65~m/s^2 and for a total time of t_1=17~s. The body is initially at rest, so the distance covered is given by
S= \frac{1}{2}a_1t_1^2=382.9~m
Calling v_f and v_i the final and initial velocity, and since the v_i=0~m/s because the body starts from rest, we can use
a= \frac{v_f-v_i}{t}
to find the final velocity after this first leg:
v_{f}=v_i+a_1t_1=45~m/s
And the average velocity in this first leg is
v_1= \frac{v_f+v_i}{2}=22.5~m/s

2) Uniform motion. The velocity is constant and it is equal to the final velocity of the first leg: v_2=45~m/s. This is also the average velocity of the second leg. 
The total time of this second leg is t_2=1.60~min = 96~s. The distance covered is given by
S_2=v_2t_2=45~m/s \cdot 96~s=4320~m

3) Uniformly decelerated motion, with constant deceleration of a_3=-9.39~m/s^2 and for a total time of t_3=4.8~s. Here, the initial velocity of the body is the final velocity of the previous leg, i.e. v_i=45~m/s. Therefore, the distance covered in this leg is given by
S_3=v_i t_3 + \frac{1}{2} a_3 t^2 =107.8~m
The final velocity in this leg is given by
v_f=v_i+at=45~m/s-9.39~m/s^2 \cdot 4.8~s = -0.07~m/s
The negative sign means that after decelerating, the body has started to go in the opposite direction. Similarly to step 1, the average velocity in this leg is given by
v_3 =  \frac{1}{2}(v_f+v_i)=  \frac{1}{2}(-0.07~m/s+45~m/s)=  22.5~m/s

4) Finally, the total distance covered in the motion is
S=S_1+S_2+S_3=382.9~m+4320~m+107.8~m=4810.7~m
To find the average velocity, we must "weigh" the average velocity of each leg for the correspondent time of that leg:
v_{ave}= \frac{v_1t_1+v_2t_2+v_3t_3}{t_1+t_2+t_3}=40.8~m/s
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Answer:

The electric field intensity is <u>30000 N/C.</u>

Explanation:

Given:

Magnitude of the point charge is, q=3\times 10^{-9}\ C

Distance of the given point from the point charge is, d=3\ cm=0.03\ m

Electric field intensity is directly proportional to the magnitude of point charge and inversely proportional to the square of the distance of the point and the given charge.

Therefore, electric field intensity 'E' at a distance of 'd' from a point charge 'q' is given as:

E=\frac{kq}{d^2}

Plug in k=9\times 10^9\ N\cdot m^2/C^2, q=3\times 10^{-9}\ C, d=0.03\ m. Solve for 'E'.

E=\frac{(9\times 10^9\ N\cdot m^2/C^2)(3\times 10^{-9}\ C)}{(0.03\ m)^2}\\\\E=\frac{27}{0.0009}\ N/C\\\\E=30000\ N/C

Therefore, the electric field intensity at a point 3 cm from the point charge is 30000 N/C.

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3 years ago
A 5 meter long ladder leans against a wall. The bottom of the ladder slides away from the wall at the constant rate of 1 3 m/s.
Oksana_A [137]

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

x^2+y^2=L^2------1

at x=3

y^2=25-9=16

y=4 m

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}

3\times 13=-4\times \frac{\mathrm{d} y}{\mathrm{d} t}

\frac{\mathrm{d} y}{\mathrm{d} t}=-\frac{3\times 13}{4}=-9.75 m/s

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Answer:

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Explanation:

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3 years ago
a 4kg metal block absorbs 5000j of energy and increases to a temperature of 22°c. the metal has a specific heat capacity of 250j
e-lub [12.9K]

Answer:

17 °C

Explanation:

From specific Heat capacity.

Q = cm(t₂-t₁)................. Equation 1

Where Q = Heat absorb by the metal block, c = specific heat capacity of the metal block, m = mass of the metal block, t₂ = final temperature, t₁ = Initial temperature.

make t₁ the subject of the equation

t₁ = t₂-(Q/cm)............... Equation 2

Given: t₂ = 22 °C, Q = 5000 J, m = 4 kg, c = 250 J/kg.°c

Substitute into equation 2

t₁ = 22-[5000/(4×250)

t₁ = 22-(5000/1000)

t₁ = 22-5

t₁ = 17 °C

6 0
3 years ago
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