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Korvikt [17]
3 years ago
11

An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F

ind (a) the magnitude and (b) the direction of the sum of the momenta of the two joggers. Describe the direction as an angle with respect to due east.
Physics
1 answer:
ololo11 [35]3 years ago
7 0

Answer

given,

mass of jogger  = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed  = 1.3 m/s  in  61 ° north of east.

jogger one

P_1 = m_1 v_1 \hat{i}

P_1 = 67 \times 2.3\hat{i}

P_1 = 154.1 \hat{i}

P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}

P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}

P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}

P_2 = 44.12\hat{i} +79.59\hat{j}

now

P = P₁ + P₂

P = 198.22 \hat{i} +79.59 \hat{j}

magnitude

P = \sqrt{198.22^2 + 79.59^2}

P =213.60 kg.m/s

\theta = tan^{-1}\dfrac{79.59}{198.22}

\theta = 21.87

the angle is \theta = 21.87 north of east

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Suppose that a comet has a very eccentric orbit that brings it quite close to the Sun at closest approach (perihelion) and beyon
Nutka1998 [239]

Answer:

16.63min

Explanation:

The question is about the period of the comet in its orbit.

To find the period you can use one of the Kepler's law:

T^2=\frac{4\pi}{GM}r^3

T: period

G: Cavendish constant = 6.67*10^-11 Nm^2 kg^2

r: average distance = 1UA = 1.5*10^11m

M: mass of the sun = 1.99*10^30 kg

By replacing you obtain:

T=\sqrt{\frac{4\pi}{GM}r^3}=\sqrt{\frac{4\pi^2}{(6.67*10^{-11}Nm^2/kg^2)(1.99*10^{30}kg)}(1.496*10^8m)^3}\\\\T=997.9s\approx16.63min

the comet takes around 16.63min

8 0
3 years ago
How do you find the total magnification of a microscope?
Alla [95]
In order to find total magnification of a microscope, you need to multiply the power of eyepiece and objective lens.

Hope this helps!
6 0
3 years ago
A ball is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s. How far will the ball go before hit
tia_tia [17]

The distance travelled by the ball that is thrown horizontally from a window that is 15.4 meters high at a speed of 3.01 m/s is 5.34 m

s = ut + 1 / 2 at²

s = Distance

u = Initial velocity

t = Time

a = Acceleration

Vertically,

s = 15.4 m

u = 0

a = 9.8 m / s²

15.4 = 0 + ( 1 / 2 * 9.8 * t² )

t² = 3.14

t = 1.77 s

Horizontally,

u = 3.01 m / s

a = 0 ( Since there is no external force )

s = ( 3.01 * 1.77 ) + 0

s = 5.34 m

Therefore, the distance travelled by the ball before hitting the ground is 5.34 m

To know more about distance travelled

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7 0
1 year ago
A force of 10 n acts on a mass of 5 kg producing an acceleration of 1.5 m/s2 what is the magnitude of the force of friction acti
prisoha [69]
The friction is 2.5N. The Net force is 10 N - 2.5 N .= 7.5 N.
acceleration = 7.5 / 5 = 1.5 m/s^2
8 0
3 years ago
A 1200.0-kg car is traveling at 19m/s. The driver suddenly slams on the brakes and skids to a stop. The coefficient of kinetic f
Alja [10]
<h2>Answer</h2>

option D)

2.4 seconds

<h2>Explanation</h2>

Given in the question,

mass of car = 1200kg

speed of car = 19m/s

Force due to direction of travel

F = ma

  = 12000(a)

Force to due frictional force in reverse direction

-F = mg(friction coefficient)

   = -12000(9.81)(0.8)

<h2>-mg(friction coefficient) = ma  </h2>

(cancelling mass from both side of equation)

g(0.8) = a

(9.81)(0.8) = a

a = 7.848 m/s²

<h2>Use Newton Law of motion</h2><h3>vf - vo = a • t</h3>

where vf = final velocity

          vo = initial velocity

          a = acceleration

           t = time

0 - 19 = 7.8(t)

t = 19/7.8

  = 2.436 s

  ≈ 2.4s

5 0
2 years ago
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