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4 years ago

7

what kind of energy does a bow string have when it's stretched? a gravitational potential energy b.elastic potential energy c.ch

emical energy d.kinetic energy

1 answer:

KIM [24]4 years ago

5 0

Hi I believe it is b. sorry if this isnt found to be helpful.

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PLEASE HELP WILL GIVE BRAINLIEST!!!!

zimovet [89]

Answer:

pelvic gridle

Explanation:

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3 years ago

5. Identify the reactants in this chemical equation:<br> 2H3PO4 → HAP20, +H2O

mixas84 [53]

Answer:

H=Hydrogen, PO4=Phosphate

Explanation:

Reactions are on the left side of arrow and products are on the right side of arrow.

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3 years ago

The water from a fire hose follows a path described by y equals 3.0 plus 0.8 x minus 0.40 x squared (units are in meters). If

ivanzaharov [21]

Explanation:

It is given that, the water from a fire hose follows a path described by equation :

$y=3+0.8x-0.4x^2$ ........(1)

The x component of constant velocity, $v_x=5\ m/s$

We need to find the resultant velocity at the point (2,3).

Let $\dfrac{dx}{dt}=v_x$ and $\dfrac{dy}{dt}=v_y$

Differentiating equation (1) wrt t as,

$\dfrac{dy}{dt}=0.8\times \dfrac{dx}{dt}-0.8x\times \dfrac{dx}{dt}$

$v_y=0.8\times v_x-0.8x\times v_x$

$v_y=0.8v_x(1-x)$

When x = 2 and $v_x=5\ m/s$

So,

$v_y=0.8\times 5\times (1-2)$

$v_y=-4\ m/s$

Resultant velocity, $v=\sqrt{v_x^2+v_y^2}$

$v=\sqrt{5^2+(-4)^2}$

v = 6.4 m/s

So, the resultant velocity at point (2,3) is 6.4 m/s. Hence, this is the required solution.

8 0

3 years ago

A 0.270 m radius, 510-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a

vlada-n [284]

Answer:

0.35701 T

Explanation:

$B_i$ = Initial magnetic field

$B_f$ = Final magnetic field

$\phi$ = Magnetic flux

t = Time taken = 4.17 ms

N = Number of turns = 510

$\epsilon$ = Induced emf = 10000 V

r = Radius = 0.27 m

A = Area = $\pi r^2$

Induced emf is given by

$\epsilon=-N\frac{d\phi}{dt}\\\Rightarrow \epsilon=-N\frac{B_fAcos90-B_iAcos0}{dt}\\\Rightarrow \epsilon=N\frac{B_iA}{dt}\\\Rightarrow B_i=\frac{\epsilon dt}{NA}\\\Rightarrow B_i=\frac{10000 \times 4.17\times 10^{-3}}{510\times \pi 0.27^2}\\\Rightarrow B_i=0.35701\ T$

The magnetic field strength needed is 0.35701 T

4 0

4 years ago

students make a small elevator machine with 5 kg and 10 kg masses on either side how fast will the masses accelerate once they a

mihalych1998 [28]

Answer:

The acceleration of the elevator machine is, a = 3 m/s²

Explanation:

Given data,

The mass on the one end, m = 5 kg

The mass on the other end, M = 10 kg

According to the Atwood's machine

Ma = Mg - T

ma = T - mg

Adding those equations,

a (M + m) = g ( M - m)

a = (M + m) / ( M - m)

Substituting the values,

a = (10 + 5) / (10 - 5)

= 3 m/s²

The acceleration of the elevator machine is, a = 3 m/s²

8 0

3 years ago