We're adding two vectors here. The first is 300 Newtons to the right, which we can write as (300, 0), meaning 300 to the right, 0 up.
The second is 300 at let's say a 45 degree angle down. For the components we have an isosceles right triangle with hypotenuse 300, so the components are both magnitude 300/√2 = 150√2. So we can write this vector (150√2, -150√2), the negative sign because it points down in the y direction.
Adding is componentwise. The resulting force is (300+150√2, -150√2).
That has square magnitude
r² = (300+150√2)² + (-150√2)² = 150² ( (2+√2)² + (√2)² )
= 150²( (6 + 4√2) + 2)
= 300²(2+√2)
so
r = 300 √(2+√2) Newtons
That's the answer; I'm not sure if your class expects a calculator approximation, which is 554.3 Newtons.
Answer:
232 J/K
Explanation:
The amount of heat gained by the air = the amount of heat lost by the tea.
q_air = -q_tea
q = -mCΔT
q = -(0.250 kg) (4184 J/kg/ºC) (20.0ºC − 85.0ºC)
q = 68,000 J
The change in entropy is:
dS = dQ/T
Since the room temperature is constant (isothermal):
ΔS = ΔQ/T
Plug in values (remember to use absolute temperature):
ΔS = (68,000 J) / (293 K)
ΔS = 232 J/K
Answer:
v = -0.45 m/s
Explanation:
Assuming the canoe was initially at rest with momentum L = 0
and that the dog's velocity is in the positive direction
conservation of momentum
0 = 15(1.2) + 40v
v = -0.45 m/s
Answer is B- 200 m
Given:
m (mass of the car) = 2000 Kg
F = -2000 N
u(initial velocity)= 20 m/s.
v(final velocity)= 0.
Now we know that
<u>F= ma</u>
Where F is the force exerted on the object
m is the mass of the object
a is the acceleration of the object
Substituting the given values
-2000 = 2000 × a
a = -1 m/s∧2
Consider the equation
<u>v=u +at</u>
where v is the initial velocity
u is the initial velocity
a is the acceleration
t is the time
0= 20 -t
t=20 secs
s = ut +1/2(at∧2)
where s is the displacement of the object
u is the initial velocity
t is the time
v is the final velocity
a is the acceleration
s= 20 ×20 +(-1×20×20)/2
<u>s= 200 m</u>
