For this case, what we can do is use the Pythagorean theorem to find the magnitude of the displacement of the car.
We have then

From here, we clear the value of d.
We have then:

Rewriting:
Answer:
The magnitude of the car's displacement is:
d = 20 miles
Answer:
335°C
Explanation:
Heat gained or lost is:
q = m C ΔT
where m is the mass, C is the specific heat capacity, and ΔT is the change in temperature.
Heat gained by the water = heat lost by the copper
mw Cw ΔTw = mc Cc ΔTc
The water and copper reach the same final temperature, so:
mw Cw (T - Tw) = mc Cc (Tc - T)
Given:
mw = 390 g
Cw = 4.186 J/g/°C
Tw = 22.6°C
mc = 248 g
Cc = 0.386 J/g/°C
T = 39.9°C
Find: Tc
(390) (4.186) (39.9 - 22.6) = (248) (0.386) (Tc - 39.9)
Tc = 335
Answer:
β2= β1+10*f
Explanation:
comparing β2 and β1, it is said that β2 is increased by a factor of f.
for each factor of f, there is a 10*f dB increase.
therefore if the β1 is increases by an intensity of factor f
the new intensity would be β1+ 10*f
Answer:
The y-component of the normal force is 45.74 N.
Explanation:
Given that,
Mass of the crate, m = 5 kg
Angle with hill, 
We need to find the y component of the normal force. We know that the y component of the normal force is given by :

So, the y-component of the normal force is 45.74 N. Hence, this is the required solution.
(a) Both the girl and the boy have the same nonzero angular displacement.
Explanation:
The angular displacement of an object moving in uniform circular motion, as the boy and the girl on the merry-go-round, is given by

where
is the angular speed
t is the time interval
For a uniform object in uniform circular motion, all the points of the object have same angular speed. This means that the value of
is the same for the boy and the girl.
Therefore, if we consider the same time interval t, the boy and the girl will also have same nonzero angular displacement.
(b) The girl has greater linear speed.
Explanation:
The linear (tangential) speed of a point along the merry-go-round is given by

where
is the angular speed
r is the distance of the point from the centre of the merry-go-round
In this problem, the girl is near the outer edge, while the boy is closer to the centre: since the value of
is the same for both, this means that the value of r is larger for the girl, so the girl will also have a greater linear speed.