Answer:
a = 30 [m/s²]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration. In this way, we have the following equation.
∑F = m*a
where:
m = mass = 2.5 [kg]
a = acceleration [m/s²]
F = force = 75 [N]
Now replacing:
[tex]75=2.5*a\\a=75/2.5\\a=30[m/s^{2} ][/tex]
11. A bowling ball has a mass of 7 kg and a velocity of 3 m/s.
(A) What is the ball's momentum?
Answer: 21 kg x m/s
Momentum can be found using the equation p = mv, or momentum = mass times velocity.
Therefore, to find the momentum of the bowling bowl, you can plug the mass and velocity values into the equation to get p = (7kg)(3m/s)
Solve the equation to get a momentum of 21kg x m/s
I hope I could help :)
HELP!!!!!
Explain how the sea star reproduce and why the fisherman returned to more sea stars. (page 29 - 32) (book in case needed: california inspire science reproduction of organisms grade 6 integrated unit 2 student edition) (Two parts for full points - 10 points for how the sea star reproduces including what it is called and 10 points to explain why the fisherman found more stars.) <----- That is part of a question not the points i can give you,
Answer:
A sea star that has the ability to regenerate amputated limbs must first undergo a repair phase to heal the exposed wound. Once the wound is healed, the sea star can begin to generate new cells, which in turn, sparks new growth. Regeneration can take anywhere from several months to years. Reproduction: Sea stars are broadcast spawners. Males release sperm into the water and females release eggs. The fertilized eggs hatch into a larval form that lives as plankton, sometimes for months, before settling on the sea floor in its adult form..
Ans: 10.0096 cm
What should be the initial lengths of steel and copper rods so that the length of steel rod is 5 cm
larger than the copper rod at all temperatures?
An iron sphere has a radius of 10 cm at a temperature of 0°C. Calculate the change in volume of the
Ans: 1.383 cm
sphere it is heated 100°. 9.3710-
Ans: 17 cm, 12 cm
Answer:
dab is the answerjdjdjnejee
Mercury is a liquid metal having a density of 13.6 g/mL. What is the
volume of 0.45 kg of mercury metal?
Answer:
[tex]\boxed {\boxed {\sf v \approx 33.088 \ mL}}[/tex]
Explanation:
The formula for density is:
[tex]d= \frac{m}{v}[/tex]
where m is the mass and v is the volume.
The mass is 0.45 kilograms and the density is 13.6 grams per milliliter. The density is given in grams, so we must convert the mass.
There are 1000 grams in 1 kilogram or [tex]\frac{1000 \ g} {1 \ kg}[/tex]. We can multiply the mass by this ratio.
[tex]0.45 \ kg * \frac{1000 \ g} {1 \ kg} = 0.45 * 1000 \ g = 450 \ g[/tex]
Now we have values for the mass and density:
[tex]13.6 \ g/mL =\frac{450 \ g}{ v}[/tex]
Cross multiply.
[tex]\frac {13.6 \ g/mL}{1} =\frac{450 \ g}{ v}[/tex]
[tex]13.6 \ g/mL * v=450 \ g * 1 \\13.6 \ g/mL * v=450 \ g[/tex]
We are trying to find the volume, so we must isolate that variable.
13.6 and v are being multiplied. The inverse of multiplication is division. Divide both sides by 13.6
[tex]\frac {13.6 \ g/mL*v } { 13.6 \ g/mL} = \frac{ 450 \ g} {13.6 \ g/mL}[/tex]
[tex]v= \frac{ 450 \ g} {13.6 \ g/mL}[/tex]
The grams will cancel.
[tex]v= 33.0882353 \ mL\\v \approx 33.088 \ mL[/tex]
The volume is about 33.088 milliliters.