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madreJ [45]
3 years ago
11

Lila is a track and field athlete.She has to complete four laps around the track, which is 400 meters. The race took her 6 minut

es to complete which best describes her speed and velocity
Physics
1 answer:
hoa [83]3 years ago
3 0
V=d/t
V=?
d=400m(4)
  =1600m
t=6 min.
 =360 s

V=1600m/360s
V=4.4m/s
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What happens is a series circuit when you increase the number of bulbs?
Volgvan

The bulbs will produce lesser light than their capacity, In short they will be dimmer because the the energy will get divided in the number of bulbs.

5 0
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An object is thrown upwards with the speed of 40.0 km/hr in 5 second, calculate the distance in millimeters (mm).
sertanlavr [38]
V=(40km/hr)(hr/3600s)(1000000mm/km)
v=11111.1mm/s

v=d/t
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4 0
3 years ago
19. A mass of gas has a volume of 4 m3, a temperature of 290 K, and an absolute pressure of 475 kPa. When the gas is allowed to
Aleks [24]

Recall this gas law:

\frac{P₁V₁}{T₁} = \frac{P₂V}{T₂}

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V₁ and V₂ are the initial and final volumes.

T₁ and T₂ are the initial and final temperatures.

Given values:

P₁ = 475kPa

V₁ = 4m³, V₂ = 6.5m³

T₁ = 290K, T₂ = 277K

Substitute the terms in the equation with the given values and solve for Pf:

\frac{475*4}{290} = \frac{P₂*6.5}{277}

<h3>P₂ = 279.2kPa</h3>
8 0
3 years ago
A baseball of mass m = 0.31 kg is spun vertically on a massless string of length L = 0.51m. The string can only support a tensio
natulia [17]

Given data:

* The mass of the baseball is 0.31 kg.

* The length of the string is 0.51 m.

* The maximum tension in the string is 7.5 N.

Solution:

The centripetal force acting on the ball at the top of the loop is,

\begin{gathered} T+mg=\frac{mv^2}{L}_{} \\ v^2=\frac{L(T+mg)}{m} \\ v=\sqrt[]{\frac{L(T+mg)}{m}} \end{gathered}

For the maximum velocity of the ball at the top of the vertical circular motion,

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where g is the acceleration due to gravity,

Substituting the known values,

\begin{gathered} v_{\max }=\sqrt[]{\frac{0.51(7.5_{}+0.31\times9.8)}{0.31}} \\ v_{\max }=\sqrt[]{\frac{0.51(10.538)}{0.31}} \\ v_{\max }=\sqrt[]{17.34} \\ v_{\max }=4.16\text{ m/s} \end{gathered}

Thus, the maximum speed of the ball at the top of the vertical circular motion is 4.16 meters per second.

8 0
1 year ago
An object of mass 'm' is on an inclined plane with an
Levart [38]

Answer:

F_{\text{par}} = F_{\text{frict}}

F_{\text{norm}} = F_{\text{perp}}

Explanation:

See attachment for complete work.

Download pdf
7 0
2 years ago
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