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Llana [10]
3 years ago
7

Problem 3.28 (a) In outer space, far from other objects, block 1 of mass 51 kg is at position <5, 15, 0> m, and block 2 of

mass 1550 kg is located at position <13, 15, 0> m. What is the (vector) gravitational force acting on block 2 due to block 1? It helps to make a sketch of the situation.
Physics
1 answer:
lbvjy [14]3 years ago
5 0

Answer: 8.242 × 10 exp -8 N

Explanation: F = G *m* M/r^2

r = √(13-5)^² + (51-15)^² + (0-0)²

r = 8; the distance between the masses

G =6.673 × 10 -¹¹ Nm²kg-²

F = gravitational force of attraction of m(51kg) on M {1550kg)

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Jimmy held the end of a metal bar over a fire while holding on to the opposite end. After a few minutes, the end he was holding
Pavel [41]

Answer:

A. Conduction

Explanation:

Conduction is the transfer of heat within the material itself, and he's holding the metal bar over the fire.

6 0
4 years ago
Three point charges are located on the x-axis. The first charge, q1 = 10 μC, is at x = -1.0 m. The second charge, q2 = 20 μC, is
victus00 [196]

Answer:

<em>3.15 N towards the positive x-axis</em>

<em></em>

Explanation:

first charge has charge q1 = 10 μC = 10 x 10^-6 C

second charge has charge q2 = 20 μC = 20 x 10^-6 C

third charge has charge q3 = -30 μC = -30 x 20^-6 C

According to coulomb's law, force between two charged particle is given as

F = \frac{-kQq}{r^2}

Where

F is the force between the charges

k is Coulomb's constant = 9 x 10^9 kg⋅m^3⋅s^−2⋅C^−2.

Q is the magnitude of one charge

q is the magnitude of the other charge

is the distance between these two charges

For the force on q2 due to q1,

distance r between them = 0 - (-1.0) = 1 m

F = \frac{-9*10^{9}*10*10^{-6}*20*10^{-6}}{1^2} = -1.8 N (the negative sign indicates a repulsion on q2 towards the positive  x-axis)

For the force on q2 due to q3,

distance between them = 2.0 - 0 = 2 m

F = \frac{-9*10^{9}*20*10^{-6}*(-30*10^{-6})}{2^2} = 1.35 N (the positive sign indicates an attraction on q2 towards the positive x-axis)

Resultant force on q2 = 1.8 N + 1.35 N = <em>3.15 N towards the positive x-axis</em>

3 0
4 years ago
A truck pushes a pile of dirt horizontally on a frictionless road with a net force of 20\, \text N20N20, start text, N, end text
meriva

Answer:

300 Nm ; 300 J

Explanation:

Given that:

Force (F) = 20 N

Distance (d) = 15 m

The kinetic energy (Workdone) = Force * Distance

Kinetic Energy = 20N * 15m

Kinetic Energy = 300Nm

K. E = 1/2

4 0
3 years ago
Read 2 more answers
Can someone help me with this please?​
Nikolay [14]

Answer:

Explanation:

umm... try 30

4 0
3 years ago
15 points. give me the method.
AveGali [126]

Answer:

\boxed{{160 \:  m(s)}^{ - 1} }

Explanation:

if \:the \:  frequencies \: are \to \\   f_{1} =  640Hz  \\ and \\f_{2}   = 480Hz \:  \\ but \:  \boxed{v = f \gamma }:   f =  \frac{v}{ \gamma } \\ if \:  \gamma_{1}  -  \gamma _{2}  = 1 =  \gamma  \\ f_{1}  - f_{2}  = 640 - 480 = \boxed{ 160Hz} = f \\ v = f \gamma = 160 \times 1 =  \boxed{{160 \:  m(s)}^{ - 1} }

5 0
3 years ago
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