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likoan [24]
3 years ago
13

All of the waves in the electromagnetic spectrum are _______ waves.

Physics
2 answers:
Scrat [10]3 years ago
7 0
The answer is B. L<span>ongitudinal.

I hope this helps. :) I just took the quiz btw!</span>
neonofarm [45]3 years ago
4 0
 I think the answer you are looking for A compression wave.
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How much heat is required to raise the temperature of 100 g of water from 5 degree celsius to 90 degree celsius?​
Scorpion4ik [409]

Explanation:

1-How many moles of NazCOs are in 10.0 ml of a 2.0 M solution?

2-How many moles of NaCl are contained in 100.0 ml of a 0.20 M solution?

3- What weight (in grams) of H2SO4 would be needed to make 750.0 ml of

2.00 M solution?

4-What volume (in ml) of 18.0 M H2SO4 is needed to contain 2.45 g H2S04?

8 0
3 years ago
An astronomer observes an asteroid in the solar system. He notes that the asteroid is three times farther from the Sun than Eart
wlad13 [49]
450 kilometers ! or 933 million miles

3 0
3 years ago
Read 2 more answers
Experiment with the battery voltage set to 15 volts, measure the current in a parallel circuit with 1,2,3, and 4 light bulbs. (I
koban [17]

Answer:

1) current = I

2) Resistance = V/I

3) current = 2I

4) resistance = V/2I

5) current = 3I

6) Resistance = V/3I

7) Current = 4I

8) Resistance = V/4I

Explanation:

When one bulb is connected across the battery then let say the current is given as I

Then resistance is given as

R = \frac{V}{I}

When two bulbs are in parallel with the battery then

total current becomes twice of initial current

so we have

current = 2I

Resistance of the circuit is now

R = \frac{V}{2I}

When three bulbs are in parallel with the battery then

total current becomes three times of initial current

so we have

current = 3I

Resistance of the circuit is now

R = \frac{V}{3I}

When four bulbs are in parallel with the battery then

total current becomes four times of initial current

so we have

current = 4I

Resistance of the circuit is now

R = \frac{V}{4I}

3 0
3 years ago
High-speed stroboscopic photographs show that the head of a 210-g golf club is traveling at 56 m/s just before it strikes a 46-g
tamaranim1 [39]

Explanation:

It is given that,

Mass of golf club, m₁ = 210 g = 0.21 kg

Initial velocity of golf club, u₁ = 56 m/s

Mass of another golf ball which is at rest, m₂ = 46 g = 0.046 kg

After the collision, the club head travels (in the same direction) at 42 m/s. We need to find the speed of the golf ball just after impact. Let it is v.

Initial momentum of golf ball, p_i=m_1u_1=0.21\ kg\times 56\ m/s=11.76\ kg-m/s

After the collision, final momentum p_f=0.21\ kg\times 42\ m/s+0.046v

Using the conservation of momentum as :

p_i=p_f

11.76\ kg-m/s=0.21\ kg\times 42\ m/s+0.046v

v = 63.91 m/s

So, the speed of the  golf ball just after impact is 63.91 m/s. Hence, this is the required solution.

3 0
3 years ago
Please I need help with this :(
Ann [662]

three charged particals are located at the corners of an equil triangle shown in the figure showing let (q 2.20 Uc) and L 0.650

3 0
2 years ago
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