Explanation: First fix your units. acceleration is in m/s/s.
T-Mg=Ma
M=348.1+169.2 =517.3 kg
T= 517.3 (9.81+4.11)
I hope this helps
The doppler radar is used in technology in two ways;
<span>·
</span>Continuous Doppler radar – it has the capability
of receiving signals in means to provide output in velocity from the target
<span>·
</span>It may be use as radar gun in which police use
to detect speeding.
The acceleration of the wagon along the ground is 3.6 m/s².
To solve the problem above, we need to use the formula of acceleration as related to force and mass.
Acceleration: This can be defined as the rate of change of velocity.
⇒ Formula:
- Fcos∅ = ma................. Equation 1
⇒ Where:
- F = Force
- ∅ = angle above the horizontal
- m = mass of the wagon
- a = acceleration of the wagon
⇒ make a the subject of equation 1
- a = Fcos∅/m..................... Equation 2
From the question,
⇒ Given:
⇒ Substitute these values into equation 2
- a = 44(cos35°)/10
- a = 44(0.8191)/10
- a = 3.6 m/s²
Hence, The acceleration of the wagon along the ground is 3.6 m/s²
Learn more about acceleration here: brainly.com/question/9408577
The statement '<span>The more particles a substance has at a given temperature, the more thermal energy it has' is true. </span><span>The
kinetic molecular theory of gases has three main laws and one of them is the
average kinetic energy of the particles in a gas. The average kinetic energy of
the gas particles is the behavior and movement it does in the surroundings. It
is directly proportional to temperature wherein if you increase the
temperature, the kinetic energy of a particle also increases. It will also
decrease its movement or its kinetic energy if the temperature lowers. </span>
Answer:
a) x = 40 t
, y = 39 t
, z = 6 + 32 t - 16 t
², b) x = 80 feet
, y = 78 feet
, the ball came into the field
Explanation:
a) This is a projectile launch exercise, where in the x and y axes there is no acceleration and in the z axis the acceleration of the acceleration of gravity, let's write the equations of motion in each axis
Since the cast is in the center of the field, let's place the coordinate system
x₀ = 0
y₀ = 0
z₀ = 6 feet
x-axis (towards end zone, GOAL zone)
x = xo + v₀ₓ t
x = 40 t
y-axis (field width)
y = y₀ + 
t
y = 39 t
z axis (vertical)
z = z₀ + v_{oz} t - ½ g t²
z = 6 + 32 t - ½ 32 t²
z = 6 + 32 t - 16 t
²
b) The player catches the ball at the same height as it came out, so we can find the time it takes to arrive
z = 6
6 = 6 + 32 t - 16 t²
(t - 2)t = 0
t=0 s
t= 2 s
The ball position
x = 40 2
x = 80 feet
y = 39 2
y = 78 feet
the dimensions of the field from the coordinate system (center of the field) are
x_total = 150 feet
y _total = 80 feet
so we can see that the ball came into the field
