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Wittaler [7]
3 years ago
12

A spring with spring constant 15 N/m hangs from the ceiling. A ball is attached to the spring and allowed to come to rest. It is

then pulled down 7.5 cm and released. The ball makes 31 oscillations in 15 s seconds.1) What is its the mass of the ball?Express your answer using two significant figures.m = _____ g2) What is its maximum speed?Express your answer using two significant figures.= _____ cm/s
Physics
1 answer:
olga nikolaevna [1]3 years ago
5 0

Explanation:

It is given that,

Spring constant of the spring, k = 15 N/m

Amplitude of the oscillation, A = 7.5 cm = 0.075 m

Number of oscillations, N = 31

Time, t = 15 s

(a) Let m is the mass of the ball. The frequency of oscillation of the spring is given by :

f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}

Total number of oscillation per unit time is called frequency of oscillation. Here, f=\dfrac{31}{15}=2.06\ Hz

m=\dfrac{k}{4\pi^2f^2}

m=\dfrac{15}{4\pi^2\times 2.06^2}

m = 0.0895 kg

or

m = 89 g

(b) The maximum speed of the ball that is given by :

v_{max}=A\times \omega

v_{max}=A\times 2\pi f

v_{max}=0.075\times 2\pi \times 2.06

v_{max}=0.970\ m/s

v_{max}=97\ cm/s

Hence, this is the required solution.

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If a bag holds 70.874 grams ,how many itemsare in the bag?​
Alina [70]

Answer:

The answer is not able to be solved, because we dont know what objects are in it, and how heavy they are. More information please!

Explanation:

7 0
4 years ago
Simple Pendulum: A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light
SIZIF [17.4K]

Answer:

The correct answer is "4.443 sec".

Explanation:

Given:

Mass of child,

= 34 kg

Mass of swing,

= 18 kg

Length,

= 4.9 m

The time period of pendulum will be:

T = 2 \pi \sqrt{4g}

  = 2 \pi \sqrt{\frac{4.9}{9.8} }

  = 4.443 \ sec  

5 0
3 years ago
Read 2 more answers
Why is the following situation impossible? A skater glides along a circular path. She defines a certain point on the circle as h
Arturiano [62]

Answer:

A skater glides along a circular path. She defines a certain point on the circle as her origin. Later on, she passes through a point at which the distance she has traveled along the path from the origin is smaller than the magnitude of her displacement vector from the origin.

So here in circular motion of the skater we can see that the total path length of the skater is along the arc of the circle while we can say that displacement is defined as the shortest distance between initial and final position of the object.

So it is not possible in any circle that arc-length is less than the chord joining the two points on the circle

As we know that arc length is given as

L = R\theta

length of chord is given as

L_c = 2Rsin(\frac{\theta}{2})

so here

L > L_c

R\theta > 2R sin(\frac{\theta}{2})

so we have

\frac{\theta}{2} > sin(\frac{\theta}{2})

6 0
3 years ago
A truck with a heavy load has a total mass of 7100 kg. It is climbing a 15∘ incline at a steady 15 m/s when, unfortunately, the
Andrej [43]

Answer:

The load has a mass of 2636.8 kg

Explanation:

Step 1 : Data given

Mass of the truck = 7100 kg

Angle = 15°

velocity = 15m/s

Acceleration = 1.5 m/s²

Mass of truck = m1 kg

Mass of load = m2 kg

Thrust from engine = T

Step 2:

⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:

T = (m1+m2)*g*sinθ

⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes  m1*gsinθ .

Resultant force on truck is F = T – m1*gsinθ  

F causes the acceleration of the truck: F= m*a

This gives the equation:

T – m1*gsinθ = m1*a  

T = m1(a + gsinθ)

Combining both equations gives:

(m1+m2)*g*sinθ = m1*(a + gsinθ)

m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ

m2*g*sinθ = m1*a

Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:

m2*g*sinθ = (7100 – m2)*a

m2*g*sinθ = 7100a – m2a

m2*gsinθ + m2*a = 7100a

m2* (gsinθ + a) = 7100a

m2 = 7100a/(gsinθ  + a)

m2 = (7100 * 1.5) / (9.8sin(15°) + 1.5)

m2 = 2636.8 kg

The load has a mass of 2636.8 kg

6 0
3 years ago
Question: Point B is 25 km due east of point A. Starting from point A, a camel walks 20 km in
Korolek [52]

<em>Resultant angle; θ = 25.59°  </em>

This question is dealing with bearings and distance.

We are told that from point A, the camel walks 20 km at 15° in the south of east direction.

Thus, d_s,e = 20 km

Resolving along the horizontal east direction gives; d_e = 20 cos 15

d_e = 19.32 km

Also, resolving along the vertical south direction gives; d_s = 20 sin 15

d_s = 5.18 km

Net vertical distance; d_vert = 8km - 5.18km = 2.72 km

Net horizontal distance; d_hor = 25km - 19.32 km = 5.68 km

Now, the resultant angle is given by;

tan θ = d_vert/d_hor

tan θ = 2.72/5.68

tan θ = 0.4789

θ = tan^(-1) 0.4789

θ = 25.59°

Read more at; brainly.com/question/22518031

8 0
3 years ago
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