A shopper pushes a grocery cart for a distance of 18m at a constant speed on level ground, against a 37.5N frictional force. he
1 answer:
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The initial speed of the shot is 15.02 m/s.
The Shot put is released at a height y<em> </em>from the ground with a speed u. It is released at an angle θ to the horizontal. In a time t, the shot put travels a distance <em>R</em> horizontally.
Pl refer to the attached diagram.
Resolve the velocity u into horizontal and vertical components, u ₓ=ucosθ and uy=u sinθ. The horizontal component remains constant in the absence of air resistance, while the vertical component varies due to the action of the gravitational force.
Write an expression for R.
Therefore,
In the time t, the net displacement of the shotput is y in the downward direction.
Use the equation of motion,
Substitute the value of t from equation (1).
Substitute -2.10 m for y, 24.77 m for R and 38.0° for θ and solve for u.
The shot put was thrown with a speed 15.02 m/s.
Answer:
b. 1.1 m
Explanation:
It is given that the total distance between the masses is equal to the length of the board, which is 3 m. Therefore,
where,
s₁ = distance of fulcrum from left mass
s₂ = distance of fulcrum from right mass
In order to achieve balance, the torque due to both masses must be equal:
s₁ = 1.1 m
Hence, the correct option is:
<u>b. 1.1 m</u>