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3241004551 [841]
3 years ago
13

Tommy runs around a track whose circumference is 400 meters. He runs a single lap in a time of 62 seconds. What is Tommy’s displ

acement
Physics
1 answer:
bija089 [108]3 years ago
8 0

Answer:

<h2>Angular Displacement 6.28 radians</h2>

Explanation:

for circular motion we are expected to solve for Angular Displacement it is measured in radian

Measurement of Angular Displacement.

we can measure it using the following relation

∅= s/r

where

s = the distance travelled by the body, and

r = radius of the circle along which it is moving.

given that

circumference c,  s= 400 m

r= ?

we have to solve for the radius

we know that circumference

c= 2\pi r

400= 2*3.142*r

400= 6.282*r

divide both sides by 6.284 we have

400/6.284

r= 63.63 m

Angular  displcament

∅= 400/63.63

∅= 6.28 radians

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An airplane has an effective wing surface area of 19.4 m2 that is generating the lift force. In level flight the air speed over
alexgriva [62]

Answer:

W =23807.68 N

Explanation:

given,

surface area of wing = 19.4 m²

speed over top wing = 67 m/s

speed under wing = 51 m/s

density of air =  1.3 kg/m³

weight of plane

From Bernoulli's principle

P_1 + \dfrac{1}{2}\rho v_1^2 = P_2 + \dfrac{1}{2}\rho_2^2

where 1 and 2 are two different locations at the same geo potential level  

so if we call 1 the lower surface and 2 the upper surface,

we find the pressure differential, P₁ -P₂

\Delta P =\dfrac{1}{2}\rho (v_2^2-v_1^2)

\Delta P =\dfrac{1}{2}\times 1.3 \times (67^2-51^2)

\Delta P =1227.2\ N/m^2

then the force acting on the plane is

F=P A

F=1227.2 x 19.4

F =23807.68 N

weight of the plane

W =23807.68 N

7 0
3 years ago
A cyclist makes the following trip along two vectors; he travels 9km to the north and then travels 6km to the east​
elena-14-01-66 [18.8K]

Answer:

Final distance from the origin: 10.82 km. the vector points as shown in the attached image.

Angle with respect to the east: 56.31^o

Explanation:

Please refer to the attached image. The cyclist's trip is indicated with the green arrows (9 km to the north followed by 6 km to the east.

So his final position is at the tip of this last vector, and indicated by the orange vector drawn form the point where the trip starts to the cyclist's final location.

We observe that this orange vector is in fact the hypotenuse of a right angle triangle, and we can estimate the distance from the origin by the Pythagorean theorem:

d=\sqrt{9^2+6^2} \\d=\sqrt{81+36} \\d=\sqrt{117} \\d=10.82 \,\,km

Notice that this is NOT the actual number of km that the cyclist pedaled to reach the final point.

Now, to find the value of the angle \theta, we need to use trigonometry, and in particular the tangent function gives us the ratio between the side of the triangle "opposite" to the angle, divided the side "adjacent" to the angle:

tan(\theta)=\frac{opp}{adj} \\tan(\theta)=\frac{9}{6}\\tan(\theta)=\frac{3}{2}\\

Now we can find the value of the angle by using the arctan function:

tan(\theta)=\frac{3}{2} \\\theta=arctan(\frac{3}{2} )\\\theta= 56.31^o

6 0
3 years ago
A pair of eyeglass frames is made of epoxy plastic. At room temperature (20.0°C), the frames have circular lens holes 2.50 cm in
nignag [31]

Answer:

T₂ = 114 °C

Explanation:

Area or superficial expansion: can be defined as an increase in area, per unit area per degree rise in temperature. It unit is (1/K) or (1/°C).

β = ΔA/(A₁ΔT) ............................................. equation 1

β = 2α ......................................................... equation 2

Area of circle = πr².................................... equation 3

Where β = Area expansivity, α = linear expansivity, ΔA = increase in area = (A₂ - A₁), ΔT = change in temperature, A₁ = initial area, A₂ = Final area r = radius,

from the question, The coefficient of linear expansion for epoxy = 1.3 × 10⁻⁴ °C⁻¹,  

∴ Coefficient of area expansion for epoxy = 2 × 1.3 × 10⁻⁴ =

β = 2.6 × 10⁻⁴ °C⁻¹,

Using equation 3 to calculate for area, and taking (π = 3.143)

r₁ = 2.5 cm ∴ A₁ = πr₁² = 3.143 × 2.5² =19.64 cm².

r₂ = 2.53 cm ∴ A₂ = πr₂² = 3.143 × 2.53² =20.12 cm².

ΔA = A₂ - A₁ = 20.12 - 19.64 = 0.48 cm².

Making ΔT the subject of the relation in equation 1.

ΔT = ΔA/(βA₁) ...................................... equation 4

Substituting the values above into equation 4,

ΔT = 0.48/(2.6 × 10⁻⁴ × 19.64)

ΔT = 0.48/(51.064 × 10⁻⁴)

ΔT =( 0.48/51.064) × 10000

ΔT = 0.0094 × 10000 = 94 °C

But, ΔT = T₂ -T₁,

Then, T₂ = ΔT  + T₁

Where T₁ = 20 °C,  ΔT =94 °C

∴ T₂  = 94 + 20 = 114 °C.

Therefore, temperature at which the frame must be heated if the the lenses 2.53 cm are to be inserted in them = 114  °C

3 0
3 years ago
2019kL is the same as__dL?
Gwar [14]
If kl is kiloliters and dl is deciliters, 2019kl=2.019e+7
4 0
3 years ago
Distance between bholu and golu house house 9 km . bholu has to attend golu birthday party at 7am . he started from his home at
navik [9.2K]
For the rest 3 km


Bholu need to run in 5 minutes so


He need to 3 × 1000 / 5 × 60 m/s

3000/ 300 = 10 m/s


So , his speed should be 1 m/s to reach in 5 minute



mark as brainliest 5 star
6 0
4 years ago
Read 2 more answers
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