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Blizzard [7]
3 years ago
14

A 3.00-kg object undergoes an acceleration given by a = (2.00 i + 5.00 j) m/s^2. Find (a) the resultant force acting on the obje

ct and (b) the magnitude of the resultant force.
Physics
1 answer:
kobusy [5.1K]3 years ago
3 0

Answer:

(a): The resultant force acting on the object are F= (5.99 i + 14.98 j).

(b): The magnitude of the resultant force are F= 16.4 N < 68.19º .

Explanation:

m= 3kg

a= 2 i + 5 j = 5 .38 < 68.19 º

F= m * a

F= 3* ( 5.38 < 68.19º )

F= 16.4 N < 68.19º

Fx= F * cos(68.19º)

Fx= 5.99

Fy= F* sin(68.19º)

Fy= 14.98

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Interest groups representing businesses and investors are often among the most successful lobbying groups in foreign policy. Why
alisha [4.7K]

The members of these groups make up the majority of voters in many districts thus this  be considered a problem.

<u>Option: D</u>

<u>Explanation:</u>

Interest groups play a key role in US politics. Such organizations are made up of wealthy and powerful members who often seek to impose some form of leverage in politicians to promote their goals and agendas. Across the years via many campaigns, they have understood how to speak and manipulate elected leaders and apply leverage to get the kind of legislation that is in their favor. Here the majority of voters in several districts are standing due to group members, as we recognize the interest group belongs to a body in which it uses different methods of lobbying to influence others.

7 0
3 years ago
If a spring has a spring constant of 1.00 × 10^3 N/m, what is the restoring force when the mass has been displaced 20.0 cm?
never [62]

The spring has a spring constant of 1.00 * 10^3 N/m and the mass has been displaced 20.0 cm then the restoring force is 20000 N/m.

Explanation:

When a spring is stretched or compressed its length changes by an amount x from its equilibrium length then the restoring force is exerted.

spring constant is k = 1.00 * 10^3 N/m

mass is x = 20.0 cm

According to Hooke's law, To find restoring force,

    F = - kx

        = - 1.00 *10 ^3 * 20.0

     F = 20000 N/m

Thus, the spring has a spring constant of 1.00 * 10^3 N/m and the mass has been displaced 20.0 cm then the restoring force is 20000 N/m.

3 0
2 years ago
Help meee please please please
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8 0
2 years ago
If a ball is rolling in a straight line and you push it to the right its velocity will to the right
Keith_Richards [23]

The answer would be "velocity will stay in the same direction." Since the ball is going in the same the ball is still going straight but just right. Many people get confused over velocity and speed, speed it the average amount on how fast a object is going and velocity is the amount of force a object has when it has speed. So, in this case since how the ball is going straight but just right it would stay in the same direction.

Hope this helps!

4 0
3 years ago
Problem: The circular blad on a radial arm saw is turning at262
kobusy [5.1K]

Answer:

Net torque, \tau=-0.033\ N-m

Explanation:

It is given that,

Initial angular speed of the blade, \omega_i=262\ rad/s

Final angular speed of the blade, \omega_f=85\ rad/s

Time, t = 18 s

Radius of the disk, r = 0.13 m

Mass of the disk, m = 0.4 kg

We need to find the net torque applied to the blade. We know that in rotational mechanics the net torque acting on an object is equal to the product of moment of inertia and the angular acceleration such that,

\tau=I\times \alpha

The moment of inertia of the disk, I=\dfrac{mr^2}{2}

\tau=\dfrac{mr^2}{2}\times \dfrac{\omega_f-\omega_i}{t}

\tau=\dfrac{0.4\times (0.13)^2}{2}\times \dfrac{85-262}{18}

\tau=-0.033\ N-m

Negative sign shows that the net torque is acting in the opposite direction of its motion. Hence, this is the required solution.

3 0
3 years ago
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