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son4ous [18]
3 years ago
10

What is the frequency of this wave 1 2 3 4

Physics
1 answer:
viva [34]3 years ago
3 0
The frequency of this wave is 3
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Which skateboarder has greater momentum?
Verizon [17]

Answer:

skateboard b

Explanation:

p=mv

skateboard a

p=(60kg)(1.5m/s)=90kg*m/s

skateboard b

p=(50kg)(2m/s)=100kg*m/s

5 0
3 years ago
Could someone please explain to me Archimede's principle without using the term displace (or at least explaining what it means)?
disa [49]
Does this help?

When an object is immersed in a fluid (in this case water, but may include both liquids and gases) the fluid exerts an upward force on the object which is called buoyancy force or <span>up-thrust.  Archimedes’ Principle states that the buoyant force (upward push or force) applied to an object is equal to the weight of the fluid that the object takes the space of by that object.  Thus when an object is placed in water the rise in the water level is dictated by the mass of that object.</span>

<span>
</span>

<span>So for example if you fill a bucket with water and you drop a stone in that bucket, if you measure the weight of the water that overflows from the bucket due to the stone being dropped into the bucket is equivalent to the pushing force that the water has on the stone (as the stone drops to the bottom of the bucket the water is pushing it to stay afloat but the rock is more dense than water and as such its downthrust exceeds water's upthrust).</span>


4 0
3 years ago
PLEASE HELP!!!
pickupchik [31]
The answer is ultraviolet rays

7 0
3 years ago
A test charge of -3x10^-7 C is located 7 cm to the right of a charge of -9x10^-6 C and 20 cm to the left of a charge of +10x10^-
musickatia [10]

Answer:

5.634 N rightwards

Explanation:

qo = - 3 x 10^-7 C

q1 = - 9 x 10^-6 C

q2 = 10 x 10^-6 C

r1 = 7 cm = 0.07 m

r2 = 20 cm = 0.2 m

The force on test charge due to q1 is F1 which is acting towards right

According to the Coulomb's law

F_{1}=\frac{Kq_{1}q_{0}}{r_{1}^{2}}

F1 = (9 x 10^9 x 9 x 10^-6 x 3 x 10^-7) / (0.07 x 0.07)

F1 = 4.959 N rightwards

The force on test charge due to q2 is F1 which is acting towards right

According to the Coulomb's law

F_{2}=\frac{Kq_{2}q_{0}}{r_{2}^{2}}

F2 = (9 x 10^9 x 10 x 10^-6 x 3 x 10^-7) / (0.2 x 0.2)

F2 = 0.675 N rightwards

Net force on the test charge

F = F1 + F2 = 4.959 + 0.675 = 5.634 N rightwards

3 0
3 years ago
Ayudaaaaa por favorrrrrrrr!
abruzzese [7]
Same for #1 I have to say the same thing
6 0
3 years ago
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