Answer:
The coefficient of static friction is : 0.36397
Explanation:
When we have a box on a ramp of angle 
, and the box is not sliding because of friction, one analyses the acting forces in a coordinate system system with an axis parallel to the incline.
In such system, the force of gravity acting down the incline is the product of the box's weight times the sine of the angle:
Recall as well that component of the box's weight that contributes to the Normal N (component perpendicular to the ramp) is given by:
and the force of static friction (f) is given as the static coefficient of friction (
) times the normal N:
When the box starts to move, we have that the force of static friction equals this component of the gravity force along the ramp:
Now we use this last equation to solve for the coefficient of static friction, recalling that the angle at which the box starts moving is 20 degrees:
<span> 4. 1 and 2 only.
1. the downward force is the force of gravity.
2. The upward force exerted is the Normal reaction from the floor.
</span>
Answer:
15 deg
Explanation:
Assume both snowballs are thrown with the same initial speed 27.2 m/s. The first snowball is thrown at an angle of 75◦ above the horizontal. At what angle should you throw the second snowball to make it hit the same point as the first? The acceleration of gravity is 9.8 m/s 2 . Answer in units of ◦ .
Given:
For first ball, θ1 = 75◦
initial velocity for both the balls, u = 27.2 m/s
for second ball, θ2 = ?
since distance covered by both the balls is same.
Therefore,..
R1=(u^{2} sin2\alpha _{1}) /g[/tex]
the range for the first ball
the range for the second ball
R2=(u^{2} sin2\alpha _{2}) /g[/tex]
(u^{2} sin2\alpha _{2}) /g[/tex]=(u^{2} sin2\alpha _{1}) /g[/tex]
sin2\alpha _{2})=sin2\alpha _{1})
=sin^-1(sin2\alpha _{1})
=1/2sin^-1(sin2\alpha _{1})
=
15 deg
Answer:
F = 5702.56 N
Explanation:
Given that,
Mass of a small car, m = 800 kg
Initial speed of the car, u = 27.8 m/s
Final speed, v = 0
Time, t = 3.9 s
We need to find the force did it take for the car to stop.
The force acting on an object is given by :
So, the magnitude of force acting on the car to stop is 5702.56 N.
Answer:
No
Explanation:
The vertical component of Jack's initial velocity is:
5.0
⋅
sin
30
∘
=
5.0
⋅
1
2
=
2.5
m/s
With gravitational acceleration
9.8
m/s
2
, he will reach the highest point of his trajectory after:
2.5
9.8
≈
0.255
s
The average vertical component of his velocity in that
0.255
s
will be:
1
2
⋅
2.5
=
1.25
m/s
So the highest point of his trajectory will be:
0.255
⋅
1.25
≈
0.32
m
So he will pass approximately
7
cm
above the top of the candle.
The horizontal component of his velocity will be a constant:
5.0
⋅
cos
30
∘
=
5.0
⋅
√
3
2
≈
4.33
m/s
So Jack's trajectory will be substantially longer than it is high and he will spend little time anywhere near above the candle.
