Answer:
acceleration, a = 9.8 m/s²
Explanation:
'A ball is dropped from the top of a building' indicates that the initial velocity of the ball is zero.
u = 0 m/s
After 2 seconds, velocity of the ball is 19.6 m/s.
t = 2s, v = 19.6 m/s
Using
v = u + at
19.6 = 0 + 2a
a = 9.8 m/s²
Answer:
a) v₃ = 19.54 km, b) 70.2º north-west
Explanation:
This is a vector exercise, the best way to solve it is finding the components of each vector and doing the addition
vector 1 moves 26 km northeast
let's use trigonometry to find its components
cos 45 = x₁ / V₁
sin 45 = y₁ / V₁
x₁ = v₁ cos 45
y₁ = v₁ sin 45
x₁ = 26 cos 45
y₁ = 26 sin 45
x₁ = 18.38 km
y₁ = 18.38 km
Vector 2 moves 45 km north
y₂ = 45 km
Unknown 3 vector
x3 =?
y3 =?
Vector Resulting 70 km north of the starting point
R_y = 70 km
we make the sum on each axis
X axis
Rₓ = x₁ + x₃
x₃ = Rₓ -x₁
x₃ = 0 - 18.38
x₃ = -18.38 km
Y Axis
R_y = y₁ + y₂ + y₃
y₃ = R_y - y₁ -y₂
y₃ = 70 -18.38 - 45
y₃ = 6.62 km
the vector of the third leg of the journey is
v₃ = (-18.38 i ^ +6.62 j^ ) km
let's use the Pythagorean theorem to find the length
v₃ = √ (18.38² + 6.62²)
v₃ = 19.54 km
to find the angle let's use trigonometry
tan θ = y₃ / x₃
θ = tan⁻¹ (y₃ / x₃)
θ = tan⁻¹ (6.62 / (- 18.38))
θ = -19.8º
with respect to the x axis, if we measure this angle from the positive side of the x axis it is
θ’= 180 -19.8
θ’= 160.19º
I mean the address is
θ’’ = 90-19.8
θ = 70.2º
70.2º north-west
Answer:
A). A virtual image cannot be formed on a screen.
Explanation:
A virtual image can not be formed on a screen.
For image:
1.A virtual image can be viewed by the unaided eye.
2. A real image must be erect or maybe inverted.
3.Mirrors can produce virtual as well as real image ,it depends on which type of mirror is.
4.A virtual image can be photographed.
So the option A is correct.
A bathroom scales works due to gravity. Under normal
conditions, a reading can be obtained when your body is pushing some force on
the scale. However in this case, since you and the scale are both moving
downwards, so your body is no longer pushing on the scale. Therefore the answer
is:
<span>The reading will drop to 0 instantly</span>
There are some missing data in the text of the problem. I've found them online:
a) coefficient of friction dry steel piston - steel cilinder: 0.3
b) coefficient of friction with oil in between the surfaces: 0.03
Solution:
a) The force F applied by the person (300 N) must be at least equal to the frictional force, given by:
where
is the coefficient of friction, while N is the normal force. So we have:
since we know that F=300 N and
, we can find N, the magnitude of the normal force:
b) The problem is identical to that of the first part; however, this time the coefficienct of friction is
due to the presence of the oil. Therefore, we have:
