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nadya68 [22]
3 years ago
15

+90

Physics
1 answer:
Taya2010 [7]3 years ago
4 0

goU 234 = 234.04095amu & mass of He4 = 4.002602 amu)

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A ball is thrown with an initial velocity of u=(10i +15j) m/s. Whan it reaches the top of it trajectory neglecting air resistanc
liraira [26]

Answer:

v = (10 i ^ + 0j ^) m / s,    a = (0i ^ - 9.8 j ^) m / s²

Explanation:

This is a missile throwing exercise.

On the x axis there is no acceleration so the velocity on the x axis is constant

           v₀ₓ =  10 m / s

On the y-axis velocity is affected by the acceleration of gravity, let's use the equation

           v_y = v_{oy} - g t

           v_{y}^2 = v_{oy}^2 - 2 g (y - y_o)

at the highest point of the trajectory the vertical speed must be zero

           v_y = 0

therefore the velocity of the body is

          v = (10 i ^ + 0j ^) m / s

the acceleration is

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5 0
2 years ago
If μs is greater than some critical value, the woman cannot start the crate moving no matter how hard she pushes. calculate this
fiasKO [112]

weight = mg acts downwards <span>
normal force = N acts upwards. 
and force F acts at an angle θ below the horizontal. 
(Let us assume that the woman pushes from the left, so F is acted towards the right, which is below the horizontal) 
so that, Frictional force, f=us*N acts towards the left 

Now we balance the forces along x and y directions: 
y direction: N = mg + F sinΘ 
x direction: us * N = F cosΘ 

We let the value of µs be equal to a value such that any F will not be able to move the crate. Then, if we increase F by an amount F', then the force pushing the crate towards the right also increases by F' cosΘ. Additionally, the frictional force f must raise by exactly this amount. 
Since f can’t exceed us*N, so the normal force must increase by F' cosΘ/us. 
Also, from the y direction equation, the normal force exceeds by F' sin Θ. 

<span>These two values must be the same, therefore:
<span>us = cot θ</span></span></span>

4 0
3 years ago
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