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goldenfox [79]
4 years ago
6

How much force does the 2.0 kg block exert on the 1.0 kg block?

Physics
1 answer:
ELEN [110]4 years ago
6 0
20N..........................
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Which of newtons laws is illustrated by a squid moving forward by shooting water out behind it? Why?
Dimas [21]
I believe its newtons 3rd law for every action there is an equal but opposite reaction since the squid is moving foward by shooting the water its pushing the squid back as its reaction. Hope this helped !
5 0
4 years ago
What is the value of y?<br>18<br>10<br>2y + 4<br>10 + 2х​
DENIUS [597]

Answer:

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4 0
4 years ago
A figure skater skates across a rink of length 50 m in 12.1 seconds. a. What is the average speed of the skater? (2 points) b. I
melamori03 [73]
(a) The skater covers a distance of S=50 m in a time of t=12.1 s, so its average speed is the ratio between the distance covered and the time taken:
v= \frac{S}{t}= \frac{50 m}{12.1 s}=4.13 m/s

(b) The initial speed of the skater is
v_i = 4 m/s
while the final speed is
v_f = 5.3 m/s
and the time taken to accelerate to this velocity is t=2 s, so the acceleration of the skater is given by
a= \frac{v_f - v_i}{t}= \frac{5.3 m/s-4.0 m/s}{2.0 s}=0.65 m/s^2

(c) The initial speed of the skater is 
v_i = 13.0 m/s
while the final speed is 
v_f=0
since she comes to a stop. The distance covered is S=8 m, so we can use the following relationship to find the acceleration of the skater:
2aS=v_f^2 -v_i^2
from which we find
a= \frac{-v_i^2}{2S}= \frac{-(13.0m/s)^2}{2 \cdot 8.0 m}=-10.6 m/s^2
where the negative sign means it is a deceleration.
4 0
4 years ago
A 1kg ball is dropped (from rest) 100m onto a spring with spring constant 125N/m. How much does the spring compress?
Anastasy [175]

Answer:

3.95 m

Explanation:

m = 1 kg, h = 100 m, k = 125 N/m

Let the spring is compressed by y.

Use the conservation of energy

potential energy of the mass is equal to the energy stored in the spring

m x g x h = 1/2 x ky^2

1 x 9.8 x 100 = 0.5 x 125 x y^2

y^2 = 15.68

y = 3.95 m

7 0
4 years ago
In Millikan's oil-drop experiment, one looks ata small oil drop held motionless between two plates. Take the voltage betwee the
AleksandrR [38]

Answer:

The charge on the drop  is

q = 1.741 x 10 ⁻²¹ C

Explanation:

Electric field due to plates

Ef = V/d  

Ef = 2033 V / (2.0 * 10^-2 m )

Ef = 101650 V/m

So, we can write  

Ef * q = m*g

q = m*g / E f

The mass can be equal using the density and the volume so:

m = ρ * v

The volume can be find as:

v = 2.298 x 10 ⁻ ¹⁶ m³

q =  ρ * v * g / Ef

q =  81 x 10 ³ kg/ m³ * 2.2298 x 10 ⁻ ¹⁶ m³ * 9.8 m/s² / 101650 V/m

The charge on the drop  is

q = 1.741 x 10 ⁻²¹ C

5 0
3 years ago
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