Method and answer please.
method and answer please. - 1

Answers

Answer 1
Answer:

Answer:

(a) 200.0628 N   ;   ( b) 4.08Kg

Explanation:

We know,

w = mg        [here w= weight , m= mass , g= gravity]

now,

gravity on earth (g) =9.807 m/s²

mass of the spanner (m)=?

weight of the spanner on earth(w) = 40N

                                              ⇒ w = 40N

                                              ⇒m g =40N

                                              ⇒m = (40/g) Kg

                                              ⇒m= (40/9.807)Kg

                                             ∴m=4.08Kg

∴mass of the spanner on earth = 4.08Kg

We know,

Mass is the amount of matter in an object . Somass will remain same both on earth and jupiter.

So,

(b)The mass of the spanner on jupiter =  4.08Kg

Now,

The mass of the spanner on jupitar (m) =  4.08Kg

Gravity on jupiter (g*) = 5.g              [5 times on earth]

                                   = 5 . (9.807)m/s²

                                   =49.035 m/s²

(a)    

weight of the spanner on jupitar(w*) = m g*

                                                            =(4.08 x 49.035) N

                                                            =200.0628 N


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Answers

I'll provide the average distance for each since the orbit of each of the planets is not a perfect circle. The distance varies (M= Miles)

Mercury= 36,000,000 M
Venus= 67,200,000 M
Earth= 93,000,000 M
Mars= 141,600,000 M
Jupiter= 483,600,000 M
Saturn = 888,200,000 M
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Neptune = 2,798,650,000
Pluto = 3,674,500,000 M

The distance of each planet from the sun varies from each other. Mercury is 58 kilometres away, Venus is 108 kilometres away while the earth is 150 kilometres away from the sun. Mars is 228 kilometres away from the sun while Jupiter is 778 kilometres away and Saturn is 1430 kilometres away from the sun. Uranus is 2870 kilometres away from the sun, while Neptune and Pluto are 4,500 kilometres and 5,910 kilometres away from the sun.

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Answers

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A person is rowing across the river with a velocity of 4.5 km/hr northward. The river is flowing eastward at 3.5 km/hr (Figure 4). What is the magnitude of her velocity ( v ) with respect to the shore?

Answers

The magnitude of her velocity with respect to the shore will be 5.70 Km/h. The change of displacement with respect to time is defined as velocity.

What is velocity?

Velocity is defined as the change in displacement with respect to time. The quantity of velocity is a vector quantity. It is a component that is time-based. It is measured in meters per second.

The given data in the problem is;

Q is the magnitude of  Velocity of the boat = 4.5 km/hr

P is the magnitude of Velocity of the river flowing = 3.5  km/hr

R is the resultant of velocity P and Q=?

θ is the angle between the two velocities = 90°

From the law of vector addition;

\rm R= √(P^2+Q^2+2PQcos\theta) \n\n \rm R= √((3.5)^2+(4.5)^2+2*  3.5 * 4.5  cos90^0) \n\n \rm R=5.70 \ m/sec

Hence the magnitude of her velocity with respect to the shore will be 5.70 Km/h.

To learn more about the velocity refer to the link;

brainly.com/question/862972

Answer: The magnitude of her velocity ( v ) with respect to the shore is 5.70 km/h.

Explanation:

Magnitude of  Velocity of the boat = Q

Magnitude of Velocity of the river flowing = P

R = Resultant velocity due to velocity of boat and velocity of river.

Applying Law of triangles of  vector addition :

R=√(P^2+Q^2+2PQCos\theta)

From the figure attached:

P = 3.5 k/h, Q = 4.5 km/h

\theta= 90^o

R=√(P^2+Q^2+2PQCos\theta)=√((3.5)^2+(4.5)^2+3.5* 4.5* cos90^o)=5.70

(Cos90^o=0),(sin 90^o=1)

\alpha =tan^(-1)(Qsin\theta)/(P+Qcos\theta)=tan^(-1)(4.5 sin 90^o)/(3.5+4.5 cos90^o)=tan^(-1)(4.5)/(3.5)=52.12^o

The magnitude of her velocity ( v ) with respect to the shore is 5.70 km/h.

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Answers


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In 3-5 sentences, use your own words to explain how electricity gets from power plants to our homes.Use the "RAP" method to answer this question:

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Answers

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Answers

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