All categories

3 years ago

A sprinter who is running a 200-m race travels the second 100 m in much less time than the first 100m because

Physics

2 answers:

chubhunter [2.5K]3 years ago

4 0

Answer:

The sprinter must overcome inertia and accelerate during the first 100 m.

Harrizon [31]3 years ago

3 0

Answer:

He is warmed up now

Explanation:

His muscles are better and stretched now

You might be interested in

A thin, circular hoop with a radius of 0.22 m is hanging from its rim on a nail. When pulled to the side and released, the hoop

Alex73 [517]

Answer:

Period of oscillation = 1.33 seconds

Explanation:

The period of oscillation is given by:

T = 2π√[I/(MgL)]

for I = 2MR² and L = R,

Given: L = 0.22m = R

T = 2π√[2R/g]

T = 2 × 3.142 Sqrt[( 2 × 0.22)/ 9.8]

T = 6.284 Sqrt(0.44/9.8)

T = 6.284 Sqrt(0.0449)

T = 6.284 × 0.2119

T = 1.33 sec

6 0

4 years ago

A mass m = 4.6 kg hangs on the end of a massless rope L = 2.16 m long. The pendulum is held horizontal and released from rest.

serious [3.7K]

ind the velocity my using the conservation of energy mgh=1/2mv^2. At the bottom, the mass is turning in a circle with tension pointing up towards the center of the circle and weight pointing down. Use the centripetal force requirement mv^2/r=T-W and solve for T. Hope this helps.

4 0

3 years ago

A planet has two small satellites in circular orbits around the planet. the first satellite has a period 18.0 hours and an orbit

puteri [66]

The two satellites orbit around the same planet, so we can use Kepler's third law, which states that the ratio between the cube of the radius of the orbit and the orbital period is constant for the two satellites:
$\frac{r_1^3}{T_1^2}= \frac{r_2^3}{T_2^2}$
where
$r_1$ is the orbital radius of the first satellite
$r_2$ is the orbital radius of the second satellite
$T_1$ is the orbital period of the first satellite
$T_2$ is the orbital period of the second satellite

If we use the data of the problem and we re-arrange the equation, we can calculate the orbital period of the second satellite:
$T_2 = \sqrt{T_1^2 ( \frac{r_2}{r_1} )^3} = \sqrt{(18.0 h)^2 ( \frac{3\cdot 10^7 m}{2 \cdot 10^7 m} )^3} = 33.1 h$

8 0

3 years ago

How much work is required to lift a 10-newton weight from 4.0 meters to 40 meters above the surface of Earth?

kumpel [21]

<h3>Answer : 360J</h3>

s = 40m - 4m = 36m

W = F × s

= 10N × 36m = 360J

<h3>A bit of explanation : </h3>

W = Work (J)

F = Force / weight (N)

s = distance (m)

6 0

3 years ago

A particular engine has a power output of 8 kW and an efficiency of 37%. If the engine expels 10994 J of thermal energy in each

VARVARA [1.3K]

Answer:

(a) 17450.8 J (B) 1.239 sec

Explanation:

We have given power output =8 KW

Heat expelled = 10994 j

Efficiency =37% = 0.37

(A) We know that efficiency $\eta =1-\frac{heat\ expelled}{heat\ input}$

$0.37=1-\frac{10994}{heat\ input}$

Heat input =17450.8 J

(B) Work done by the engine is = 17450.8-10994=6456.8 j

Power output is given by 8 KW =8000 W

So time for each cycle is $\frac{8000}{6456.8}=1.239 sec$

3 0

3 years ago