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REY [17]
3 years ago
5

At the surface of Jupiter's moon Io, the acceleration due to gravity is 1.81m/s2 . A watermelon has a weight of 58.0N at the sur

face of the earth. In this problem, use 9.81m/s2 for the acceleration due to gravity on earth.Part AWhat is its mass on the earth's surface?Part BWhat is its mass on the surface of Io?Part CWhat is its weight on the surface of Io?
Physics
1 answer:
kvasek [131]3 years ago
6 0

Answer: A) mass on earth surface = 5.91kg

B) mass on surface of jupiter = 5.91kg

C) weight on surface of jupiter = 10.697N

Explanation:

The relationship between weight (W), mass (m) and acceleration due gravity (g) is given below

W=mg

From the question, g= 9.8m/s² and weight on the surface on the earth is 58N

A) The mass of watermelon on earth is

m = 58/ 9.8 = 5.91kg

B) the mass of the watermelon on jupiter is 5.91kg.

You will notice this is the same as the mass of watermelon on earth and that is so because mass is a scalar quantity that does not depends on the distance away from the center of the earth (unlike weight which is a vector) thus making it constant all through any location.

C) mass of watermelon is 5.91kg, g=9.8m/s² weight of watermelon on jupiter is given below as

W = mg

W = 5.91 x 9.8

= 10.697N.

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The 8.00-cm long second hand on a watch rotates smoothly.
dolphi86 [110]
The watch hand covers an angular displacement of 2π radians in 60 seconds.

ω = 2π/60
ω = 0.1 rad/s

v = ωr
v = 0.1 x 0.08
v = 8 x 10⁻³ m/s
4 0
3 years ago
A car is moving at 19 m/s along a curve on a horizontal plane with radius of curvature 49m.
JulsSmile [24]

Answer:

\mu =0.75

Explanation:

<u>Frictional Force </u>

When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

F_c=m.a_c

The centripetal acceleration a_c is computed as

\displaystyle a_c=\frac{v^2}{r}

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

F_c=m.\frac{v^2}{r}

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

F_r=\mu N

The normal force N is equal to the weight of the car, thus

F_r=\mu .m.g

Equating both forces

\displaystyle \mu .m.g=m.\frac{v^2}{r}

Simplifying

\displaystyle \mu =\frac{v^2}{rg}

Substituting the values

\displaystyle \mu =\frac{19^2}{(49)(9.8)}

\boxed{\mu =0.75}

7 0
4 years ago
You're driving your new sports car at 80 mph over the top of a hill that has a radius of curvature of 540 m. What fraction of yo
luda_lava [24]

Answer:

75.84%

Explanation:

We were given Speed of the sports car, v as 80 mph , we can convert to m/s for unit consistency.

v=80mph= 35.76 m/s

The radius of curvature is given as , r = 540 m

✓ the normal weight can be denoted as Wn

✓ the apparent weight of the person can be denoted as Wa

Wn= normal weight= mg

Wa=apparent weight = (mg - mv^2/r)

g= acceleration due to gravity= 9.8m/s^2

The apparent weightand normal weight has a ratio of

Mn/Ma= [mg - mv^2/r]/mg ........eqn(1)

If we simplify eqn(1) we have

Mn/Ma=[g - vr^2/g].............eqn(2)

Then substitute the given values

Mn/Ma=9.8 - [(35.76^2)/540]/ 9.8

=0.758×100%

Mn/Ma=75.84%

Hence, the required fraction is 75.84%

4 0
3 years ago
Projectile motion in two dimensions cannot be determined by breaking the problem into two connected one-dimensional problems.
andriy [413]
The answer I believe is False
8 0
3 years ago
The volume of an ideal gas is adiabatically reduced from 151 L to 80.6 L. The initial pressure and temperature are 1.50 atm and
Zolol [24]

Answer:

gas is dioatomic

T_f = 330.0 K

\eta = 7.07 mole

Explanation:

Part 1

below equation is used to determine the type Gas by determining \gamma value

\frac{V_{1}}{V_{F}}\gamma=\frac{P_{i}}{P_{f}}

where V_i and V_f is initial and final volume respectively

and P_i and P_f are initial and final pressure

\gamma = \frac{ln(P_f/P_i)}{ln(V_i/V_f)}

\gamma = \frac{ln(3.61/1.50)}{ln(151/80.6}

\gamma = 1.38

therefore gas is dioatomic

Part 2

final temperature in adiabatic process is given as

T_f = T_i*[\frac{v_i}{V_f}](^\gamma-1)

substituing value to get final temperature

T_f = 260*[\frac{151}{80.6}]^ {(1.38-1)}

T_f = 330.0 K

Part 3

determine number of moles by using following formula

\eta =\frac{PV}{RT}

\eta =\frac{1.013*10^{5}*0.151}{8.314*260}

\eta = 7.07 mole

4 0
3 years ago
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