<span>A measurement
that both magnitude and direction is a vector quantity. An example of this is a
moving car. The car exerts force due to its thrust and weight that runs in it. This
will give us the magnitude of the car. The resulting motion of the car in terms
of displacement, velocity and acceleration that determines its direction makes
it a vector quantity. On the other hand, a measurement that has only magnitude is
a scalar quantity. The energy exerted by the engine of the car is a scalar
quantity.</span>
Answer: 12Mg/h
Explanation:
Let the spring is compressed by a distance x,before the lift stops,then
Mg(h+x)= 1/2 kx^2 ............... 1
Kx - Mg = M ( 5g ) ............ 2
Make x the subject in equation 2
Kx = 5Mg + Mg
Kx = 6Mg
x = 6Mg/k ............ 3
Put equation 3 into 1
Mg ( h + x ) = 1/2 kx^2
Mgh + Mgx = 1/2kx^2
Mgh + Mg × 6Mg/k = 1/2k × ( 6Mg/k )^2
Mgh + Mg× 6Mg/k = 1/2k 36M^2g^2/ k^2
h =18Mg/k - 6Mg/h
K = 12Mg/h
Answer: A wave with a frequency of 14 Hz has a wavelength of 3 meters. At what speed will this wave travel? 1. = 3m (4. = 42m. 2. ... 1,7m (46) = 7802 m. 4. A wave traveling at 230 m/sec has a wavelength of 2.1 meters. What is the frequency of.
Explanation: please give me brainlest
Answer:
The lose of thermal energy is, Q = 22500 J
Explanation:
Given data,
The mass of aluminium block, m = 1.0 kg
The initial temperature of block, T = 50° C
The final temperature of the block, T' = 25° C
The change in temperature, ΔT = 50° C - 25° C
= 25° C
The specific heat capacity of aluminium, c = 900 J/kg°C
The formula for thermal energy,
<em>Q = mcΔT</em>
= 1.0 x 900 x 25
= 22500 J
Hence, the lose of thermal energy is, Q = 22500 J
Answer:

Explanation:
M = Mass of each star
T = Time period = 15.5 days
v = Orbital velocity = 230 km/s
G = Gravitational constant = 
Radius of orbit is given by

We have the relation

The mass of each star is 