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Nina [5.8K]
2 years ago
10

A motorcycle has a constant acceleration of 3.49 m/s2. Both the velocity and acceleration of the motorcycle point in the same di

rection. How much time is required for the motorcycle to change its speed from (a)29.0 to 39.0 m/s, and (b)59.0 to 69.0 m/s?
Physics
1 answer:
Vilka [71]2 years ago
7 0

Answer:

(a)2.865 s

(b)2.865 s

Explanation:

We are given that

Acceleration,a=3.49 m/s^2

a.Initial speed,u=29 m/s

Final speed,v=39 m/s

We know that

t=\frac{v-u}{a}

Using the formula

t=\frac{39-29}{3.49}=2.865 s

b.Initial speed,u=59 m/s

Final speed,v=69 m/s

Again using the formula

t=\frac{69-59}{3.49}=2.865 s

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A 15.0-m uniform ladder weighing 500 N rests against a frictionless wall. The ladder makes a 60.08 angle with the horizontal. (a
grigory [225]

Answer:

a)  fr = 266.92 N,   fy = 1300 N,  b)    μ = 0.36

Explanation:

a) This is a balancing act.

Let's write the rotational equilibrium relations, where the turning point is the bottom of the ladder and the counterclockwise rotations are positive

             -w x - W x₂ + R y = 0         (1)

usemso trigonometry to find distances

            cos 60.08 = x / 7.5

            x = 7.5 cos 60.08

            x = 3.74 m

fireman

           cos 60.08 = x₂ / 4

           x2 = 4 cos 60

           x2 = 2 m

wall support

           sin 60.08 = y / 15

           y = 15 are 60.08

           y = 13 m

we substitute in equation 1

           R y = w x + W x2

            R = (w x + W x2) / y

            R = (500 3.74 +800 2) / 13

            R = 266.92 N

now let's write the expressions for the translational equilibrium

X axis

           R -fr = 0

           R = fr

           fr = 266.92 N

Y Axis  

           Fy - w-W = 0

           fy = 500 + 800

           fy = 1300 N

b) ask the friction coefficient

the firefighter's distance is

          cos 60.08 = x₃ / 9.00

          x₃ = 9 cos 60

          x₃ = 5.28 m

from equation 1

          R = (w x + W x₃) / y

          R = 500 3.74 + 800 5.28) / 13

          R = 468.769 N

we saw that

          fr = R = 468.769

The expression for the friction force is

          fr = μ N

in this case the normal is the ratio to pesos

        N = Fy

       N = 1300 N

        μ N = fr

        μ = fr / N

        μ = 468,769 / 1300

         μ = 0.36

7 0
2 years ago
A gas sample occupies 4.2 L at a pressure of 101 kPa. What volume will it occupy if the pressure is increased to 235 kPa?
Dmitrij [34]
For this we want to use Boyle's Law. Boyle’s law states that the pressure and volume of a fixed quantity of a gas are inversely proportional under constant temperature conditions. The formula for this is P1V1 = P2V2. We want to solve this out so it equals V2 (Volume 2). So P1V1 / P2 = V2. Then plug in your values for the variables. So (101)(4.2) / 235 = V2; so 424.2 / 235 = V2. The final volume equals 1.81. I hope this helps, If not I am very sorry.
7 0
3 years ago
Read 2 more answers
A tourist averaged 82km/hr for a 6.5h trip in her volkswagen. how far did she go?
Whitepunk [10]
82 ÷ 6.5 = 12.615384615384... repeating
round = 12.6
12.6 · 6.5 = 81.9
round = 82
She went an average of 12.6 km an hour
Hope this helps! ;)
4 0
2 years ago
An automobile is traveling away from Jill and towards Jack. The horn is honking, producing a sound wave.
wlad13 [49]

Answer:

a. The sound will travel at the speed to both Jill and Jack

b. Jack

Explanation:

a. Doppler effect describes how the frequency of a sound wave changes with regards to an observer that has relative motion to the sound source;

The Doppler effect is given by the following formula;

For a sound that is moving away, as observed by Jill, we have;

f' = \dfrac{(v - v_0)}{(v + v_s)} \cdot f =  \dfrac{v }{(v + v_s)} \cdot f

For approaching sound, as observed by Jack, we have;

f' = \dfrac{(v + v_0)}{(v - v_s)} \cdot f =  \dfrac{v }{(v - v_s)} \cdot f

Where;

f = The sound wave's actual frequency

f' = The frequency of the moving sound to the observer

v = The speed of the sound wave

v₀ = The observer's velocity = 0

v_s = The velocity of the source (the automobile honking) of the sound wave

From the above equation, we have that the speed of sound, 'v', is the same to both the source, and the observer although the frequency, and therefore, the wavelength of the sound alternatively increases or decreases

b. From the Doppler effect equation, the person who will hear the highest frequency is given by the formula for the frequency when the sound is approaching the observer, which has the lower denominator

Therefore, given that the automobile is travelling towards Jack, jack will hear the higher frequency

5 0
3 years ago
How deep can an object with 6360N hitting on a sponge get ???
borishaifa [10]

Answer:

The sponge must go \dfrac{636}{m}\ \text{meter} deep

Explanation:

If F = 6360 N, then it is required to find how deep can an object with this force hitting on a sponge get.

We know that, F = mgh

m is mass

g is acceleration due to gravity

h=\dfrac{F}{mg}\\\\h=\dfrac{6360}{10m}\\\\h=\dfrac{636}{m}\ \text{meter}

So, the sponge must go \dfrac{636}{m}\ \text{meter} deep.

5 0
3 years ago
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