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Anna35 [415]
3 years ago
13

Twovectors Å and °B are such that A = 2and A-B-C find the angle between them​

Physics
1 answer:
sweet-ann [11.9K]3 years ago
7 0

Answer:

hjg567 A 567

Explanation:

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A 15 W electric shaver is used for 3 minutes. How much energy does it use in joules?​
borishaifa [10]
P=change in E/t
Change in E=p*t
=15*3
=45

The answer is 45J.
3 0
2 years ago
NEED HELP NOW WILL GIVE BRAINLIEST
Arturiano [62]

Answer:

B. solar flares sunspots,  solar flares prominences

Explanation:

Some features of the Sun's surface include sunspots, solar flares, and prominences

pls mark me brainliest its the right answer

6 0
2 years ago
Read 2 more answers
Humans must depend on the environment for their survival.<br> True<br> False
Alisiya [41]

Answer:

true

Explanation:

with out trees we would have no air with out animals the envirounment would slowly decay

8 0
2 years ago
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Two football players collide head-on in midair while trying to catch a thrown football. The first player is 98.5 kg and has an i
romanna [79]

Answer:

v_f=0.825m/s

Explanation:

We must use conservation of linear momentum before and after the collision, p_i=p_f

Before the collision we have:

p_i=p_1+p_2=m_1v_1+m_2v_2

where these are the masses are initial velocities of both players.

After the collision we have:

p_f=(m_1+m_2)v_f

since they clong together, acting as one body.

This means we have:

m_1v_1+m_2v_2=(m_1+m_2)v_f

Or:

v_f=\frac{m_1v_1+m_2v_2}{m_1+m_2}

Which for our values is:

v_f=\frac{(98.5kg)(6.05m/s)+(119kg)(-3.5m/s)}{(98.5kg)+(119kg)}=0.825m/s

7 0
3 years ago
The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the cente
Gennadij [26K]

Answer:

18.1 × 10⁻⁶ A = 18.1 μA

Explanation:

The current I in the wire is I = ∫∫J(r)rdrdθ

Since J(r) = Br, in the cylindrical wire. With width of 10.0 μm, dr = 10.0 μm. r = 1.20 mm. We have a differential current dI. We integrate first by integrating dθ from θ = 0 to θ = 2π.

So, dI = J(r)rdrdθ

dI/dr = ∫J(r)rdθ = ∫Br²dθ = Br²∫dθ = 2πBr²

Now I = (dI/dr)dr at r = 1.20 mm = 1.20 × 10⁻³ m and dr = 10.0 μm = 0.010 mm = 0.010 × 10⁻³ m

I = (2πBr²)dr = 2π × 2.00 × 10⁵ A/m³ × (1.20 × 10⁻³ m)² × 0.010 × 10⁻³ m  =  0.181 × 10⁻⁴ A = 18.1 × 10⁻⁶ A = 18.1 μA

3 0
3 years ago
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