2*3.5 = 7m/s
You multiply the acceleration per the time (they both are in seconds, otherwise, you should set them in the same units).
Complete Question:
Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bury itself just below the surface. What would have to be the mass of this asteroid, in terms of the earth’s mass M, for the day to become 25.0% longer than it presently is as a result of the collision? Assume that the asteroid is very small compared to the earth and that the earth is uniform throughout.
Answer:
m = 0.001 M
For the whole process check the following page: https://www.slader.com/discussion/question/suppose-that-an-asteroid-traveling-straight-toward-the-center-of-the-earth-were-to-collide-with-our/
Answer:
155.38424 K
2.2721 kg/m³
Explanation:
= Pressure at reservoir = 10 atm
= Temperature at reservoir = 300 K
= Pressure at exit = 1 atm
= Temperature at exit
= Mass-specific gas constant = 287 J/kgK
= Specific heat ratio = 1.4 for air
For isentropic flow
The temperature of the flow at the exit is 155.38424 K
From the ideal equation density is given by
The density of the flow at the exit is 2.2721 kg/m³
<span>θ=0.3sin(4t)
w=0.3cost(4t)(4)=1.2cost(4t)
a=-4.8sin(4t)
cos4t max will always be 1 (refer to cos graph), for same reason, sin4t will always be 0
therefore, wmax=1.2rad/s
vAmax=r*w=250*1.2=300mm/s
(may be different if your picture/radius is from a different picture)
adt=a*r=200*-4.8sin(4t)=0 (sin(4t)=0)
adn=r*w^2=200*1.2^2=288
ad= square root of adt^2+adn^2 = 288mm/s^2</span>
Answer:
514 cal
Explanation:
In order to calculate the lost heat by the amount of water you first take into account the following formula:
(1)
Q: heat lost by the amount of water = ?
m: mass of the water
c: specific heat of water = 1cal/g°C
T2: final temperature of water = 11°C
T1: initial temperature = 12°C
The amount of water is calculated by using the information about the density of water (1g/ml):
Then, you replace the values of all parameters in the equation (1):
The amount of water losses a heat of 514 cal
