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natita [175]
3 years ago
12

Because they have an extended structure, elements that are ___________ tend to be in the ___________ state at room temperature.

Physics
1 answer:
Verdich [7]3 years ago
5 0

Answer:

metal

solid

Explanation:

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At the circus, a 100.-kilogram clown is fired at 15 meters per second from a 500.- kilogram cannon. What is the recoil speed of
photoshop1234 [79]

The recoil velocity of cannon is (4) 5.0 m/s

Explanation:

We can find the recoil velocity from the law of conservation of momentum.

The recoil velocity is velocity of body 2 after release of body 1, i.e. velocity of cannon after release of clown.

Let v2 be cannon's velocity, v1 be clown's velocity given = 15 m/sec

m1 be clown's mass = 100kg and m2 be cannon's mass given = 500kg.

So recoil velocity of cannon v2 is given by,

v2 = -(m1÷m2)v1

v2 = -(100÷500)15

v2 = -5 m/s

where the minus sign refers to the direction of cannon's recoil velocity being opposite to that of clown.

Hence, option (4)5.0 m/s is the correct answer.

6 0
3 years ago
Read 2 more answers
An athlete is running a 5-km race at an average speed of 8 km/h. She has to stop to tie her shoe. After tying her shoe, she cont
Ray Of Light [21]
The total time she took to complete the race increased
5 0
3 years ago
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
olga nikolaevna [1]

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.

5 0
3 years ago
Anyone here like conspiracies? I personally do, and I want to know what you guys think of the Last Thursday conspiracy. This is
alexdok [17]

Answer:

nope dont agree with that i think it would b a lot harder to do on a mass scale like that

Explanation:

Only way to do that is if aliens with far superior technology wise came to earth and did it

6 0
3 years ago
Two children are riding on the edge of a merry-go-round that has a mass of 100.kg and radius of 1.60m and is rotating at 20.0rpm
Gre4nikov [31]

Here since both children and merry go round is our system and there is no torque acting on this system

So we will use angular momentum conservation in this

I_1\omega_1 = I_2\omega_2

now here we have

I_1 = \frac{MR^2}{2} + m_1R^2 + m_2R^2

I_1 = \frac{100(1.60)^2}{2} + (22 + 28)(1.60)^2

I_1 = 256

Now when children come to the position of half radius

then we will have

I_2 = \frac{MR^2}{2} + m_1(\frac{R}{2})^2 + m_2(\frac{R}{2})^2

I_2 = \frac{100(1.6)^2}{2} + (28 + 22)(0.8)^2

I_2 = 160

now from above equation we have

256 (20.0 rpm) = 160(\omega_2)

\omega_2 = 32 rpm

8 0
3 years ago
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