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3 years ago

Why will a change in velocity have a greater affect on KE than a change in mass?

Physics

2 answers:

Sever21 [200]3 years ago

5 0

Answer: the answer is corect of what everone else said

Explanation:

lys-0071 [83]3 years ago

5 0

Answer:

..: Check website

[02/06, 7:25 PM] Nabhaan: Check website

[02/06, 7:25 PM] Nabhaan: Check website[02/06, 6:21 PM] DHARUN ARULSELVAN: Check website

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[02/06, 7:22 PM] DHARUN ARULSELVAN: Check website

[02/06, 7:22 PM] DHARUN ARULSELVAN: Check website[02/06, 7:25 PM] Nabhaan: Check website

[02/06, 7:25 PM] Nabhaan: Check website

[02/06, 7:25 PM] Nabhaan: Check website[02/06, 6:21 PM] DHARUN ARULSELVAN: Check website

[02/06, 7:16 PM] DHARUN ARULSELVAN: Check website

[02/06, 7:22 PM] DHARUN ARULSELVAN: Check website

[02/06, 7:22 PM] DHARUN ARULSELVAN: Check website[02/06, 7:25 PM] Nabhaan: Check website

[02/06, 7:25 PM] Nabhaan: Check website

[02/06, 7:25 PM] Nabhaan: Check website[02/06, 6:21 PM] DHARUN ARULSELVAN: Check website

[02/06, 7:16 PM] DHARUN ARULSELVAN: Check website

[02/06, 7:22 PM] DHARUN ARULSELVAN: Check website

[02/06, 7:22 PM] DHARUN ARULSELVAN: Check website[02/06, 7:25 PM] Nabhaan: Check website

[02/06, 7:25 PM] Nabhaan: Check website

[02/06, 7:25 PM] Nabhaan: Check website[02/06, 6:21 PM] DHARUN ARULSELVAN: Check website

[02/06, 7:16 PM] DHARUN ARULSELVAN: Check website

[02/06, 7:22 PM] DHARUN ARULSELVAN: Check website

[02/06, 7:22 PM] DHARUN ARULSELVAN: Check website[02/06, 7:25 PM] Nabhaan: Check website

[02/06, 7:25 PM] Nabhaan: Check website

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How frequently do most coastal areas experience high tides?

Alex777 [14]

It doesn't matter where you are. "Tides" happen twice (hi - lo - hi - lo) in about 24 hours and 50 minutes. Anywhere.

If it doesn't do two complete cycles in 24 hours 50 minutes, then it's not tide.

7 0

2 years ago

Determine the magnitude of the force between two 42 m-long parallel wires separated by 0.03 m, both carrying 6.3 A in the same d

xz_007 [3.2K]

Answer:

The magnitude of the force between the two parallel wires is 0.0111 N.

Explanation:

Given;

length of the two parallel wires, L = 42 m

distance between the two wires, r = 0.03 m

current in both wires, I₁, I₂ = 6.3 A

Therefore, the magnitude of the repulsive force between the two parallel wires is given by;

$F = \frac{\mu_0 I_1I_2l}{2\pi r}\\\\where;\\\mu_0 \ is \ permeability \ of \ free \ space = 4\pi *10^{-7} \ T.m/A \\\\F = \frac{(4\pi *10^{-7})(6.3)^2(42)}{2\pi (0.03)}\\\\F = 0.0111 \ N$

Therefore, the magnitude of the force between the two parallel wires is 0.0111 N.

6 0

3 years ago

Your code returns a number of 99.123456789 +0.00455679 for your calculation. How should you report it in your lab write-up?

Aneli [31]

Answer: Your code returns a number of 99.123456789 +0.00455679

Ok, you must see where the error starts to affect your number.

In this case, is in the third decimal.

So you will write 99.123 +- 0.004 da da da.

But you must round your results. In the number you can see that after the 3 comes a 4, so the 3 stays as it is.

in the error, after the 4 comes a 5, so it rounds up.

So the final presentation will be 99.123 +- 0.005

you are discarding all the other decimals because the error "domains" them.

6 0

3 years ago

Question 30

stellarik [79]

Answer: $0.69\°$

Explanation:

The angular diameter $\delta$ of a spherical object is given by the following formula:

$\delta=2 sin^{-1}(\frac{d}{2D})$

Where:

$d=16 m$ is the actual diameter

$D=1338 m$ is the distance to the spherical object

Hence:

$\delta=2 sin^{-1}(\frac{16 m}{2(1338 m)})$

$\delta=0.685\° \approx 0.69\°$ This is the angular diameter

3 0

3 years ago

In each of the parts of this question, a nucleus undergoes a nuclear decay. Determine the resulting nucleus in each case.

vaieri [72.5K]

Answer:

(A). The resulting nucleus is Fr.

(B). The resulting nucleus is Po.

(C). The resulting nucleus is Ne

(D). The resulting nucleus is Tc.

Explanation:

Given that,

A nucleus undergoes a nuclear decay.

(A). In alpha decay,

We know that,

When the nucleus emit alpha particle then atomic mass of particle reduce by 4 and atomic number reduce by 2.

We need to calculate the resulting nucleus

Using given data

$^{227}_{89}Ac\Rightarrow ^{227-4}_{89-2}X$

$^{227}_{89}Ac\Rightarrow ^{223}_{87}Fr$

The resulting nucleus is Fr.

(B). In beta-minus decay,

We know that,

When the nucleus emit beta- minus particle then atomic mass of particle is same and atomic number increase by 1.

We need to calculate the resulting nucleus

Using given data

$^{211}_{83}Bi\Rightarrow ^{211}_{83+1}X$

$^{211}_{83}Bi\Rightarrow ^{211}_{84}Po$

The resulting nucleus is Po.

(C). In beta-plus decay,

We know that,

When the nucleus emit beta- plus particle then atomic mass of particle is same and atomic number decrease by 1.

We need to calculate the resulting nucleus

Using given data

$^{22}_{11}Na\Rightarrow ^{22}_{11-1}X$

$^{22}_{11}Na\Rightarrow ^{22}_{10}Ne$

The resulting nucleus is Ne.

(D). In gamma decay,

We know that,

When the nucleus emit gamma particle then atomic mass and atomic number of particle is same.

We need to calculate the resulting nucleus

Using given data

$^{98}_{43}Tc\Rightarrow ^{98}_{43}Tc$

The resulting nucleus is Tc.

Hence, (A). The resulting nucleus is Fr.

(B). The resulting nucleus is Po.

(C). The resulting nucleus is Ne

(D). The resulting nucleus is Tc.

4 0

3 years ago