Who invented the atomic battery and what purpose does it serve
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Magnitude of the force of tension: 139 N
Explanation:
The surface of the ramp here is assumed to be the positive x-direction.
To solve this problem and find the magnitude of the force of tension, we have to analyze only the situation along the x-direction, since the force of tension lie in this direction.
There are three forces acting along the x-direction:
- The force of tension,
, acting up along the plane
- The force of friction,
, acting down along the plane
- The component of the weight in the x-direction,
, acting down along the plane
We know that the magnitude of the weight is
So its x-component is
The net force along the x-direction can be written as
And therefore, since the net force is 98 N, we can find the magnitude of the force of tension:
Learn more about inclined planes:
#LearnwithBrainly
She threw the marshmallow at a speed of around 4.76 m/s.The formula for the horizontal range gives the velocity.
<h3>What is projectile motion?</h3>The motion of an item hurled or projected into the air, subject only to gravity's acceleration, is known as projectile motion.
The item is known as a projectile, and the course it takes is known as a trajectory. Falling object motion is a simple one-dimensional kind of projectile motion with no horizontal movement.
Given data;
The marshmallow was thrown at a distance of 2 meters
Range,R = 3 m
Initial velocity,u
The angle at which the marshmallow was thrown,θ = 30°
The acceleration due to gravity,g = 9.81 m/s²
The projectile's motion is divided into two parts: horizontal and vertical motion.
Hence, she throws the marshmallow at a speed of 4.76 m/sec.
To learn more about the projectile motion refer to the link;
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Answer:
a) v = 16.57 m / s, b) a = 19.6 m / s², d) N = 1.76 10³ N, N / W = 3
Explanation:
This exercise looks interesting, but I think you have some problem with the writing, the questions seem a bit disconnected from the initial text.
Let's answer the questions.
a) For this part we can use energy considerations.
Starting point. The upper part of the trajectory indicates that the arm is horizontally
Em₀ = U = m g h
in this case h = r
Final point. For lower of the trajectory
Em_f = K = ½ m v²
as they indicate that there is no friction
Em₀ = em_f
mgh = ½ m v²
v =
let's calculate
v =
v = 16.57 m / s
b) the centripetal acceleration has the formula
a = v² / r
a = 16.57² / 14.0
a = 19.6 m / s²
c) see attached where the diagram is
where N is the normal and w the weight
d) let's use Newton's second law
N-W = m a
N - mg = m ar
N = m (g + a)
let's calculate
N = 60.0 (9.8 + 19.6)
N = 1.76 10³ N
the relationship with weight is
N / W = 1.76 10³/( 60 9.8)
N / W = 3
normal is three times greater than body weight
e) the answer is reasonable since by Newton's first law the body must continue in a straight line, therefore to change its trajectory a force must be applied to deflect it
Answer:
a) f ’’ = f₀ , b) Δf = 2 f₀
Explanation:
a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.
Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source
f ’= fo
This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.
f ’’ = f’
where c represents the sound velocity in stationary blood
therefore the received frequency is
f ’’ = f₀
let's simplify the expression
f ’’ = f₀ \frac{c+v}{c-v}
f ’’ = f₀
b) At the low speed limit v <c, we can expand the quantity
(1 -x)ⁿ = 1 - x + n (n-1) x² + ...
f ’’ = fo
f ’’ = fo
leave the linear term
f ’’ = f₀ + f₀ 2
the sound difference
f ’’ -f₀ = 2f₀ v/c
Δf = 2 f₀