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Alinara [238K]
3 years ago
13

in the same viewing window. Compare the magnitudes of f (x) and g(x) when x is close to 0. Use the comparison to write a short p

aragraph explaining why
Physics
1 answer:
Ivanshal [37]3 years ago
6 0

Answer:

Due to inertia of restttttttrestrestrestrest

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What is the weight, in pounds, of a 195-kg object on jupiter?
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429.901 is the weight of 195 kg on Jupiter
6 0
3 years ago
Read 2 more answers
A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th
ZanzabumX [31]

Answer:

Part a)

v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}

Part b)

v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

Mv_o = M v_1 + m v_2

here we also use angular momentum conservation

so we have

M v_o d = M v_1 d + \beta mL^2 \omega

also we know that the collision is elastic collision so we have

v_o = (v_2 + d\omega) - v_1

so we have

\omega = \frac{v_o + v_1 - v_2}{d}

also we know

M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})

also we know

v_1 = v_o - \frac{m}{M}v_2

so we have

M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})

mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}

now we have

(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}

v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}

Part b)

Now we know that speed of the ball after collision is given as

v_1 = v_o - \frac{m}{M}v_2

so it is given as

v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})

3 0
3 years ago
A cake is removed from a 350◦F oven and placed on a cooling rack in a 70◦F room. After 30 minutes the cake is 200◦F. When will i
galben [10]

Answer:

350 F to 100 F it take approx 87.33 min  

Explanation:

given data

oven = 350◦F

cooling rack = 70◦F

time = 30 min

cake = 200◦F

solution

we apply here Newtons law of cooling  

\frac{dT}{dt} = -k(T-Ta)

\frac{dy}{dt} = \frac{d}{dt} (T(t) -Ta)

= \frac{dT}{dt} -\frac{dTa}{dt} =\frac{dT}{dt} = -k(T-Ta)

-ky \frac{dy}{dt} = -ky

T(t) -Ta = (To -Ta) e^{-kt} T(t) = Ta+ (To -Ta)  e^{-kt}

put her value for time 30 min and T(t) = 200◦F and To =350◦F  and Ta = 70◦F

so here

200 = 70 + ( 350 - 70 ) e^{-k30}

k = 0.025575

so here for  T(t) = 100F

100 = 70 + ( 350 - 70 ) e^{-0.025575*t}

time = 87.33 min

so here 350 F to 100 F it take approx 87.33 min  

5 0
3 years ago
Does the Digestive System deconstruct both mechanically and<br> chemically?
SIZIF [17.4K]

Answer:

Chemically

Explanation:

Your stomach acid would break down food and theYour stomach acid would break down food and that would be chemical.

7 0
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In a tug of war between Mrs. Brenneman and Mr. Schroedl seems at a standstill. Then Mrs. Brenneman tugs hard giving a force of 4
sergiy2304 [10]
He feels a 10 N to the left force moves. Yes ,he moves.
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3 years ago
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