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3 years ago

Find the sum of the vectors: 40m/s2 Northeast, 10 m/s2 Northeast

Physics

1 answer:

podryga [215]3 years ago

8 0

Answer:

Explanation:

Since both vectors are pointing on the same direction (Northeast), the sum of them will point in that same direction, and its magnitud will be the sum of the magnitudes of each vector (40m/s2+10m/s2). This problem is just a problem in one dimension. The sum of the vectors is then 50m/s2 Northeast.

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Answer:

B.1.898^27 kg

Explanation:

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If a girl is standing still and holding a box, is she doing any work? Why or why not?

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3 years ago

Read 2 more answers

The magnitude of each force is 208 N the force on the right is applied at an angle 36° and the mass of the block is 17 kg the co

djyliett [7]

Answer:

Explanation:

According to Newtons second law of motion

$\sm F_x = ma_x\\F_m - F_f = ma_x\\mgsin \theta - \mu R mgcos \theta = ma_x\\$

Given

Mass m = 17kg

Fm = 208N

theta = 36 degrees

g = 9.8m/s²

a is the acceleration

Substitute

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3 years ago

A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.

iris [78.8K]

To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as

$F = \frac{kq_1q_2}{d^2}$

Here

k = Coulomb's Constant

$q_{1,2} =$ Charge of each object

d = Distance

Our values are given as,

$q_1 = 1 \mu C$

$q_2 = 6 \mu C$

d = 1 m

$k = 9*10^9 Nm^2/C^2$

a) The electric force on charge $q_2$ is

$F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}$

$F_{12} = 54 mN$

Force is positive i.e. repulsive

b) As the force exerted on $q_2$ will be equal to that act on $q_1$ ,

$F_{21} = F_{12}$

$F_{21} = 54 mN$

Force is positive i.e. repulsive

c) If $q_2 = -6 \mu C$ , a negative sign will be introduced into the expression above i.e.

$F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}$

$F_{12} = F_{21} = -54 mN$

Force is negative i.e. attractive

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3 years ago

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guajiro [1.7K]

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