Plz help me with this activity :(

3) A school bus traveling at 12 m/s has a momentum of 144,000 kg.m/s.

spin [16.1K]

Answer:

does this have picture so I can solve

6 0

3 years ago

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A portable music player, operating with four 1.5 V cells connected in series, provides a resistance of 15 000 Ω. What amount of

nata0808 [166]

Answer:

Explanation:

Given that,

A portable music player is operating with 4 cell batteries connected in series, and each cell has a P.D of 1.5V.

Then,

Total potential difference is

P.D_total = V1 + V2 + V3 + V4

P.D_total = 1.5 + 1.5 + 1.5 + 1.4

P.D_total = 6V.

The music player provides a resistance of 15,000Ω

R = 15,000Ω

We want to find the current (I) flowing through the music player?

Using ohms law

V = IR

Where

V is the potential difference

I is the current

R is the resistance

Therefore,

I = V/R

I = 6 / 15,000

I = 4 × 10^-4 A

I = 0.4 × 10^-3 A

I = 0.4 mA.

So, 0.4mA is passing through the music player

7 0

3 years ago

The voltage across the secondary winding is 10,000 V, and the voltage across the primary winding is 25,000 V. If the primary win

Allisa [31]

Answer

80 coils

Explanation

The turn rule of a transformer says that, the amount of voltage induced in the secondary coil is proportional to the number of coil. It is simplified by the ratio:

Np/Ns = Vp/Vs

200/Ns = 25,000/10,000

200 × 10,000 = 25,000Ns

2,000,000 = 25000Ns

Dividing both sides by 25,000,

Ns = 2,000,000/25,000

= 80 coils

7 0

3 years ago

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Tammy leaves the office, drives 26 km due

fredd [130]

Answer:

72.98 km

Explanation:

Her displacement is simply the distance from her final position to her initial position.

Now, I've drawn and attached a triangle diagram to depict this her movement.

Point O is her initial starting point.

Point A is the first point she gets to after travelling north while point B is the final point after travelling north east.

From the triangle, the displacement will be the distance OB which is denoted by x and can be solved from cosine rule.

Thus;

x² = 62² + 26² - 2(62 × 26)cos 120

x² = 4520 + 806

x² = 5326

x = √5326

x = 72.98 km

4 0

3 years ago

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated

Lelu [443]

Incomplete question as we have not told to find what quantity.The complete question is here

A spherical capacitor contains a charge of 3.50 nC when connected to a potential difference of 210.0 V. Its plates are separated by vacuum and the inner radius of the outer shell is 5.00 cm.calculate: (a) the capacitance; (b) the radius of the inner sphere; (c) the electric field just outside the surface of the inner sphere.

Answer:

(a) $C=16.7pF$

(b) $r_{a} =3.749cm$

(c) $E=2.24*10^{4} N/C$

Explanation:

Given data

$Q=3.50nC\\V=210V\\r_{b}=5.0cm$

For part (a)

The Capacitance given by:

$C=\frac{Q}{V}\\ C=\frac{3.50*10^{-9} C}{210V}\\C=1.6666*10^{-11}F\\or\\C=16.7pF$

For part (b)

The Capacitance of coordinates is given as

$C=\frac{4\pi e}{\frac{1}{r_{a} }-\frac{1}{r_{b} } }\\ So\\{\frac{1}{r_{a} }-\frac{1}{r_{b} } }=\frac{4\pi *8.85*10^{-12} }{1.666*10^{-11}}=6.672m^{-1} \\ \frac{1}{r_{a} }=6.672+(1 /0.05)\\\frac{1}{r_{a} }=26.672\\r_{a} =1/26.672\\r_{a} =0.0375m\\r_{a} =3.749cm$

For part (c)

The electric field according to Gauss Law is given by:

$EA=\frac{Q}{e}\\ E=\frac{Q}{4\pi er_{a}^{2} }=\frac{kQ}{r_{a}^{2}}\\ E=\frac{9*10^{9}*3.50*10^{-9} }{(0.0375m)^{2} }\\ E=2.24*10^{4} N/C$

7 0

3 years ago