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DochEvi [55]
3 years ago
9

Playing tetherball, a boy hits the 0.45 kg ball, moving toward him initially at 4.6 m/s, in the opposite direction at 2.5 m/s. H

is hand contacts the ball for 0.025 seconds. What was the average net force of his hit?
Physics
1 answer:
saul85 [17]3 years ago
8 0

Answer:

HI

Explanation:

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Finding the spring constant as shown, spring 3, which has an unknown spring constant k3, replaces spring 2. the mass of the weig
nevsk [136]
Replaces spring 2. the mass of the weight and pulley are unchanged: m=5.8 kg and mp=1.7 kg
6 0
3 years ago
X(????) = 5.0???? 2 − 4.0???? 3 m.
Burka [1]

Answer:

JA

Explanation:

s of time, (b) the velocity and acceleration at t = 2.0 s, (c) the time at which the position is a maximum, (d) the time at which the velocity is zero, and (e) the maximum position. Assume all variable and constants are in SI units.

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3 years ago
Suppose a car approaches a hill and has an initial speed of
kvv77 [185]

Answer:

a) 1.73*10^5 J

b) 3645 N

Explanation:

106 km/h = 106 * 1000/3600 = 29.4 m/s

If KE = PE, then

mgh = 1/2mv²

gh = 1/2v²

h = v²/2g

h = 29.4² / 2 * 9.81

h = 864.36 / 19.62

h = 44.06 m

Loss of energy = mgΔh

E = 780 * 9.81 * (44.06 - 21.5)

E = 7651.8 * 22.56

E = 172624.6 J

Thus, the amount if energy lost is 1.73*10^5 J

Work done = Force * distance

Force = work done / distance

Force = 172624.6 / (21.5/sin27°)

Force = 172624.6 / 47.36

Force = 3645 N

5 0
3 years ago
Psuedopod in a sentence
mixas84 [53]
Does it have to be that exact word. cause it is just another term for psuedopodium

5 0
3 years ago
Add the vectors:
Anettt [7]

Vector 1 has components

x_1=(10\,\mathrm m)\cos20^\circ\approx9.40\,\mathrm m

y_1=(10\,\mathrm m)\sin20^\circ\approx3.42\,\mathrm m

and vector 2 has

x_2=(10\,\mathrm m)\cos80^\circ\approx1.74\,\mathrm m

y_2=(10\,\mathrm m)\sin80^\circ\approx9.85\,\mathrm m

Add these vectors to get the resultant, which has components

x_{\rm total}\approx11.133\,\mathrm m

y_{\rm total}\approx13.268\,\mathrm m

The magnitude of the resultant is

\sqrt{{x_{\rm total}}^2+{y_{\rm total}}^2}\approx17.321\,\mathrm m

with direction \theta such that

\tan\theta=\dfrac{y_{\rm total}}{x_{\rm total}}\implies\theta\approx50^\circ

or about 50º N of E.

8 0
3 years ago
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