Answer:
Option E is correct 310N
Explanation:
Given that the force used to push the crate is F = 200N
The force directed 20° below the horizontal
Mass of crate is m = 25kg
Weight of the crate can be determine using
W = mg
g is gravitational constant =9.8m/s²
W = 25×9.8
W = 245 N
Check attachment. For free body diagram and better understanding
Using newton second law along the vertical axis since we want to find the normal force
ΣFy = m•ay
ay = 0, since the body is not moving in the vertical or y direction
N—W—F•Sin20 = 0
N = W+F•Sin20
N = 245+ 200Sin20
N = 245 + 68.4
N = 313.4 N
The normal force is approximately 310 N to the nearest ten
The speed of the ball is 27.4 m/s
Explanation:
The speed of an object is given by:
where
d is the distance covered
t is the time taken
In this problem, we have:
d = 300 yards is the distance covered by the golf ball
t = 10 s is the time taken
Keeping in mind that
1 yard = 0.914 m
We can convert the distance from yards to meters:
And substituting into the equation, we find the speed of the ball:
Learn more about speed:
brainly.com/question/8893949
#LearnwithBrainly
You asked a question. I'm about to answer it.
Sadly, I can almost guarantee that you won't understand the solution.
This realization grieves me, but there is little I can do to change it.
My explanation will be the best of which I'm capable.
Here are the Physics facts I'll use in the solution:
-- "Apparent magnitude" means how bright the star appears to us.
-- "Absolute magnitude" means the how bright the star WOULD appear
if it were located 32.6 light years from us (10 parsecs).
-- A change of 5 magnitudes means a 100 times change in brightness,
so each magnitude means brightness is multiplied or divided by ⁵√100 .
That's about 2.512... .
-- Increasing magnitude means dimmer.
Decreasing magnitude means brighter.
+5 is 10 magnitudes dimmer than -5 .
-- Apparent brightness is inversely proportional to the square
of the distance from the source (just like gravity, sound, and
the force between charges).
That's all the Physics. The rest of the solution is just arithmetic.
____________________________________________________
-- The star in the question would appear M(-5) at a distance of
32.6 light years.
-- It actually appears as a M(+5). That's 10 magnitudes dimmer than M(-5),
because of being farther away than 32.6 light years.
-- 10 magnitudes dimmer is ( ⁵√100)⁻¹⁰ = (100)^(-2) .
-- But brightness varies as the inverse square of distance,
so that exponent is (negative double) the ratio of the distances,
and the actual distance to the star is
(32.6) · (100)^(1) light years
= (32.6) · (100) light years
= approx. 3,260 light years . (roughly 1,000 parsecs)
I'll have to confess that I haven't done one of these calculations
in over 50 years, and I'm not really that confident in my result.
If somebody's health or safety depended on it, or the success of
a space mission, then I'd be strongly recommending that you get
a second opinion.
But, quite frankly, I do feel that mine is worth the 5 points.
