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lesantik [10]
3 years ago
10

I'm looking for the answers for 4-6 and how to do them. I have some answers already but I'm very unsure of them. Thank you!

Physics
1 answer:
coldgirl [10]3 years ago
7 0
Answer to 4: E

Answer to 6: D
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The radii of the sprocket assemblies and the wheel of the bicycle in the figure are:
Furkat [3]
To solve this task we have to make a proportion, but firstly we have to set up all the main points : so, the distance is  s=r(B), that has its <span>r=radius,B=angle in rad velocity v=ds/dt= w(r)
Do not forget about </span> w = angular speed in rad/s and w1 = 1 revolution/sec = 2Pi (rad/s)
Now we can go to proportion
v1=v2
w1*r1 = w2r2w2 = w1 * r1/r2 = 2w1 = 4Pi (rad/s)
w2 = w3 (which is the   angular velocity of the rear wheel) &#10;
SOLVING FOR A : v3 = w3 * r3 = 4pi * 14 (inch/s) = 14.66 ft/sec
v3 = 14.66 ft/sec(1 mile/5280 ft)( 3600 sec/h)= 9.99 or something about <span>10 mph --- SOLVING FOR B.
</span>I'm sure it helps!
7 0
3 years ago
Please answer this. Science 7th grade.
nikdorinn [45]

Answer:

the answer would be 2

Explanation:

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6 0
3 years ago
A hose directs a horizontal jet of water, moving with a velocity of 20m/s, on to a vertical wall. The
just olya [345]
Force is defined as the rate of change of momentum.
The initial amount of momentum is mv because water stops when it hit the wall total change of momentum must be \Delta p=mv.
Now let's calculate the force.
F= \frac{dp}{dt}=\frac{d(mv)}{dt}=\frac{dm}{dt}v
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\frac{dm}{dt}=\rho Av
Our final formula would be:
F=\rho Avv=\rho Av^2
And now we can calculate the answer:
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6 0
3 years ago
If a car increases its velocity from zero to 60 m/s in 10 seconds, its acceleration is
andrezito [222]
We know that a=vf_vi/t equals equation "a" . Where a is the acceleration of the body , vf is the final velocity , vi is the initial velocity and t is equal to time . Since vi equals o m/s , vf equals to 60 m/s and t equals 10 s. Put in equation "a". a=60-0/10 =6m/s2
5 0
3 years ago
Estimate the magnitude of the electric field due to the proton in a hydrogen atom at a distance of 5.29 x 10-11 m, the expected
V125BC [204]
Would be A 1012 N/C because The magnitude of the electric field at distance r from a point charge q is E=k
e
​
q/r
2
, so
E=
(5.11×10
−11
m)
2

(8.99×10
9
N.m
2
/C
2
)(1.60×10
−19
C)
​

=5.51×10
11
N/C∼10
1
2N/C
making (e) the best choice for this question.
3 0
3 years ago
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