Smaller the surface area greater the gravitational force and vice–versa yes or no? justify the statement.
1 answer:
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Answer:
Explanation:
<u>Friction Force</u>
When objects are in contact with other objects or rough surfaces, the friction forces appear when we try to move them with respect to each other. The friction forces always have a direction opposite to the intended motion, i.e. if the object is pushed to the right, the friction force is exerted to the left.
There are two blocks, one of 400 kg on a horizontal surface and other of 100 kg on top of it tied to a vertical wall by a string. If we try to push the first block, it will not move freely, because two friction forces appear: one exerted by the surface and the other exerted by the contact between both blocks. Let's call them Fr1 and Fr2 respectively. The block 2 is attached to the wall by a string, so it won't simply move with the block 1.
Please find the free body diagrams in the figure provided below.
The equilibrium condition for the mass 1 is
The mass m1 is being pushed by the force Fa so that slipping with the mass m2 barely occurs, thus the system is not moving, and a=0. Solving for Fa
The mass 2 is tried to be pushed to the right by the friction force Fr2 between them, but the string keeps it fixed in position with the tension T. The equation in the horizontal axis is
The friction forces are computed by
Recall N1 is the reaction of the surface on mass m1 which holds a total mass of m1+m2.
Replacing in [1]
Simplifying
Plugging in the values
Answer:
Explanation:
Given that:
Charge (q) on the particle = 3 × 10⁻⁸ C
mass (m) of the particle = 6 × 10⁻⁹ kg
at a distance x = 15 cm , the velocity in the plate = 900 m/s²
For the square plate, the surface charged density σ = -8 × 10⁻⁶ C/m²
To start with calculating the electric field as a result of the square plate; we use the formula;
On the square plate; The electric force F = Eq
The acceleration
For the particle, the velocity at distance x = 7 m can be calculated by using the formula:
From the calculation, we realize that the charge acting between the particle and the plate is said to be "opposite".
Hence, the force is an attractive force.
Similarly, there is a gradual increase exhibited by the velocity of the particle.
Therefore, the particles get to the detector, but the detector failed to get detect due to the velocity which is greater than 1000 m/s.