The picture isn’t clear so I can’t read the dimensions of the box but I can try my best to guide u through the question.
For part a u need to find the volume of the box as that will equal the volume of sand that can be filled inside.
For this u multiply the height, width and length of the box.
For part b the mass of sand alone will be
=Mass of box + sand - Mass of empty box
=216 - 40
=176 grams
For part c the density of sand can be calculated by the formula
Density= Mass/Volume
So the mass (176g) / volume from part a
For part d u need to know that something will float if it has a lower density than what it is floating in. If the final density of sand that was found in part c is less than the density of gold (19.3 g/cm^3) it will float. Otherwise it will sink.
Hope this helped!
The acceleration of the car will be needed in order to calculate the time. It is important to consider that the final speed is equal to zero:
We can clear time in the speed equation:
If you find some mistake in my English, please tell me know.
Answer:
It's the 3rd option
Explanation:
Wind is caused by the differences in air pressure on Earth's surface.
Answer:
largest lead = 3 m
Explanation:
Basically, this problem is about what is the largest possible distance anchorman for team B can have over the anchorman for team A when the final leg started that anchorman for team A won the race. This show that anchorman for team A must have higher velocity than anchorman for team B to won the race as at the starting of final leg team B runner leads the team A runner.
So, first we need to calculate the velocities of both the anchorman
given data:
Distance = d = 100 m
Time arrival for A = 9.8 s
Time arrival for B = 10.1 s
Velocity of anchorman A = D / Time arrival for A
=100/ 9.8 = 10.2 m/s
Velocity of anchorman B = D / Time arrival for B
=100/10.1 = 9.9 m/s
As speed of anchorman A is greater than anchorman B. So, anchorman A complete the race first than anchorman B. So, anchorman B covered lower distance than anchorman A. So to calculate the covered distance during time 9.8 s for B runner, we use
d = vt
= 9.9 x 9.8 = 97 m
So, during the same time interval, anchorman A covered 100 m distance which is greater than anchorman B distance which is 97 m.
largest lead = 100 - 97 = 3 m
So if his lead no more than 3 m anchorman A win the race.
Answer: The bottom of the ladder is moving at 3.464ft/sec
Explanation:
The question defines a right angle triangle. Therefore using pythagorean
h^2 + l^2 = 10^2 = 100 ...eq1
dh/dt = -2ft/sec
dl/ dt = ?
Taking derivatives of time in eq 1 on both sides
2hdh/dt + 2ldl/dt = 0 ....eq2
Putting l = 5ft in eq2
h^ + 5^2 = 100
h^2 = 25 = 100
h Sqrt(75)
h = 8.66 ft
Put h = 8.66ft in eq2
2 × 8.66 × (-2) + 2 ×5 dl/dt
dl/dt = 17.32 / 5
dl/dt = 3.464ft/sec
