The acceleration of the sled will be 1.30 m/s². Force is defined as the product of mass and acceleration.
<h3>What is force?</h3>
Force is defined as the push or pulls applied to the body. Sometimes it is used to change the shape, size, and direction of the body.
Given data;
m(mass of sled)=8 kg
Θ is the inclination of force= 50°
Force of friction,f=2.4 N.
The applied force at the given angle is resolved into the two-component as;
The net vertical force is zero;
From Newton's second law the net force as;
Hence, the acceleration of the sled will be 1.30 m/s².
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Answer:
time required after impact for a puck is 2.18 seconds
Explanation:
given data
mass = 30 g = 0.03 kg
diameter = 100 mm = 0.1 m
thick = 0.1 mm = 1 × m
dynamic viscosity = 1.75 × Ns/m²
air temperature = 15°C
to find out
time required after impact for a puck to lose 10%
solution
we know velocity varies here 0 to v
we consider here initial velocity = v
so final velocity = 0.9v
so change in velocity is du = v
and clearance dy = h
and shear stress acting on surface is here express as
= µ
so
= µ ............1
put here value
= 1.75× ×
= 0.175 v
and
area between air and puck is given by
Area =
area =
area = 7.85 × m²
so
force on puck is express as
Force = × area
force = 0.175 v × 7.85 ×
force = 1.374 × v
and now apply newton second law
force = mass × acceleration
- force =
- 1.374 × v =
t =
time = 2.18
so time required after impact for a puck is 2.18 seconds
<span>Electrons display some properties of waves and while they reside outside of the nucleus, their positions cannot be known with certainty. </span>
Answer:
a) q = 39.29 cm
, b) h ’= - 3.929 cm the image is inverted and REAL
Explanation:
For this exercise we will use the equation of the constructor
1 / f = 1 / p + 1 / q
where f is the focal length of the salad bowl, p and q are the distance to the object and the image
The metal salad bowl behaves like a mirror, so its focal length is
f = R / 2
f = 44/2
f = 22 cm
a) Suppose that the distance to the object is p = 50 cm, let's find the distance to the image
1 / q = 1 / f - 1 / p
1 / q = 1/22 - 1/50
1 / q = 0.0254
q = 39.29 cm
b) to calculate the size of the image we use the equation of magnification
m = h’/ h = - q / p
h ’= - q / p h
h ’= - 39.29 / 50 5
h ’= - 3.929 cm
the negative sign means that the image is inverted
as the rays of light pass through the image this is REAL