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3 years ago

Match the labels to the correct location on the roller coaster.

Physics

2 answers:

Step2247 [10]3 years ago

7 0

Just put the answers that have the same letter near them to the letters in the boxes

Pachacha [2.7K]3 years ago

6 0

At the bottom left,

Increasing potential energy.

At ascending,

Increasing kinetic energy.

At the peak,

Greatest potential energy.

At the descending,

Greatest kinetic energy.

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An ideal spring with spring constant k is hung from the ceiling. The initial length of the spring, with nothing attached to the

hram777 [196]

The mass m of the object = 5.25 kg

<h3>Further explanation</h3>

Given

k = spring constant = 3.5 N/cm

Δx= 30 cm - 15 cm = 15 cm

Required

the mass m

Solution

F=m.g

Hooke's Law

F = k.Δx

$\tt m.g=k.\Delta x\\\\m.10=3.5\times 15\\\\m=5.25~kg$

7 0

3 years ago

A 0.18-kg turntable of radius 0.32 m spins about a vertical axis through its center. A constant rotational acceleration causes t

borishaifa [10]

Answer:

Angular acceleration will be $18.84rad/sec^2$

Explanation:

We have given that mass m = 0.18 kg

Radius r = 0.32 m

Initial angular velocity $\omega _i=0rev/sec$

And final angular velocity $\omega _f=24rev/sec$

Time is given as t = 8 sec

From equation of motion

We know that $\omega _f=\omega _i+\alpha t$

$24=0+\alpha \times 8$

$\alpha =3rev/sec^2=3\times 2\times \pi rad/sec^2=18.84rad/sec^2$

So angular acceleration will be $18.84rad/sec^2$

4 0

4 years ago

Will give correct answer brainliest<br><br>5 kg m/s<br>8kg m/s<br>80 kg m/s<br>200 kg m/s

o-na [289]

Answer: Here this will help you..

Explanation:

1 kg-m/s to kilogram-force meter/second = 1 kilogram-force meter/second

5 kg-m/s to kilogram-force meter/second = 5 kilogram-force meter/second

10 kg-m/s to kilogram-force meter/second = 10 kilogram-force meter/second

20 kg-m/s to kilogram-force meter/second = 20 kilogram-force meter/second

30 kg-m/s to kilogram-force meter/second = 30 kilogram-force meter/second

40 kg-m/s to kilogram-force meter/second = 40 kilogram-force meter/second

50 kg-m/s to kilogram-force meter/second = 50 kilogram-force meter/second

75 kg-m/s to kilogram-force meter/second = 75 kilogram-force meter/second

100 kg-m/s to kilogram-force meter/second = 100 kilogram-force meter/second

8 0

3 years ago

A 27-g steel-jacketed bullet is fired with a velocity of 640 m/s toward a steel plate and ricochets along path CD with a velocit

Dmitry [639]

Answer:

F = - 3.56*10⁵ N

Explanation:

To attempt this question, we use the formula for the relationship between momentum and the amount of movement.

I = F t = Δp

Next, we try to find the time that the average speed in the contact is constant (v = 600m / s), so we say

v = d / t

t = d / v

Given that

m = 26 g = 26 10⁻³ kg

d = 50 mm = 50 10⁻³ m

t = d/v

t = 50 10⁻³ / 600

t = 8.33 10⁻⁵ s

F t = m v - m v₀

This is so, because the bullet bounces the speed sign after the crash is negative

F = m (v-vo) / t

F = 26*10⁻³ (-500 - 640) / 8.33*10⁻⁵

F = - 3.56*10⁵ N

The negative sign is as a result of the force exerted against the bullet

6 0

3 years ago

3. A concrete highway is built of slabs 12 m long (20o C). How wide should the expansion cracks between the slabs be (at 20o C)

lesantik [10]

Answer:

$\Delta x=2.304\times 10^{-2}\ m=2.304\ cm$ is the minimum gap between the slabs

Explanation:

Given:

length of the concrete highway, $l=12\ m$

coefficient of thermal expansion, $\alpha=12\times 10^{-6}\ ^{\circ}C^{-1}$

range of temperature variation, $(-30^{\circ}C\ to\ 50^{\circ}C) \Rightarrow \Delta T=80^{\circ}C$

<u>Now form the equation of thermal expansion:</u>

$\Delta l=l\times \alpha\times \Delta T$

$\Delta l=12\times (12\times 10^{-6})\times 80$

$\Delta l=1.152\times 10^{-2}\ m$

Since each slab of the highway expands by the above length so the minimum gap between the slabs to prevent buckling:

$\Delta x=2\times \Delta l$

$\Delta x=2\times( 1.152\times 10^{-2})$

$\Delta x=2.304\times 10^{-2}\ m=2.304\ cm$ is the minimum gap between the slabs

4 0

3 years ago