You would die you would be 209.48 Fahrenheit and being higher than 107 Fahrenheit is very bad so imagine what would happen to you :(
Answer:
a) Team A will win.
b) The losing team will accelerate towards the middle line with 0.01 m/
Explanation:
Given that Team-A pulls with a force ,
and Team-B pulls with a force ,
∵
The rope will move in the direction of force .
∴ Team-A will win.
b) Considering both the teams as one system of total mass ,
Net force on the system , = 50-45 = 5N
Applying Newtons first law to the system ,
F = ma , where 'a' is the acceleration of the system.
Since , both the teams are connected by the same rope , their acceleration would be the same.
∴ 5 = 499×a
∴ a = 0.01 m/
Answer:
Tension= (g=acceleration of gravity)
Explanation:
Given that,
A 5Kg and 10Kg are attached by a cable suspended over a pulley.
As 10Kg > 5Kg , the 10 kg mass accelerates down and the 5kg mass accelerates up, let it be a. Let the tension in the cable be T.
So, the equations of motion are
Now adding them we get,
Substituting them back in the equation we get,
Answer:
distance between school and home is 21 miles
Explanation:
given data
in rush hour speed s1 = 28 mph
less traffic speed s2 = 42 mph
time t = 1 hr 15 min = 1.25 hr
to find out
distance d
solution
we consider here distance home to school is d and t1 time to reach at school
we get here distance equation when we go home to school that is
distance = 28 × t1 .......................1
and when we go school to home distance will be
distance = 42 × ( t - t1 )
distance = 42 × ( 1.25 - t1 ) ...................2
so from equation 1 and 2
28 × t1 = 42 × ( 1.25 - t1 )
t1 = 0.75
so
from equation 1
distance = 28 × t1
distance = 28 × 0.75
distance = 21 miles
Answer:
θ = 12.95º
Explanation:
For this exercise it is best to separate the process into two parts, one where they collide and another where the system moves altar the maximum height
Let's start by finding the speed of the bar plus clay ball system, using amount of momentum
The mass of the bar (M = 0.080 kg) and the mass of the clay ball (m = 0.015 kg) with speed (v₀ = 2.0 m / s)
Initial before the crash
p₀ = m v₀
Final after the crash before starting the movement
= (m + M) v
p₀ =
m v₀ = (m + M) v
v = v₀ m / (m + M)
v = 2.0 0.015 / (0.015 +0.080)
v = 0.316 m / s
With this speed the clay plus bar system comes out, let's use the concept of conservation of mechanical energy
Lower
Em₀ = K = ½ (m + M) v²
Higher
= U = (m + M) g y
Em₀ =
½ (m + M) v² = (m + M) g y
y = ½ v² / g
y = ½ 0.316² / 9.8
y = 0.00509 m
Let's look for the angle the height from the pivot point is
L = 0.40 / 2 = 0.20 cm
The distance that went up is
y = L - L cos θ
cos θ = (L-y) / L
θ = cos⁻¹ (L-y) / L
θ = cos⁻¹-1 ((0.20 - 0.00509) /0.20)
θ = 12.95º