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sergejj [24]
3 years ago
8

A 600g basketball, a 500g football, a 150g baseball, and a 50g yo-yo: four toys. If you had to throw one of the four, which one

would take the MOST force to throw?
A) baseball
B) basketball
C) football
D) yo-yo
Physics
2 answers:
Paha777 [63]3 years ago
6 0
The basket ball, because it is heavier than the rest and hence would require more force.
Aleksandr-060686 [28]3 years ago
6 0

Answer:

The correct answer is option B.

Explanation:

Force is defined as product of mass of an object and acceleration by which object is moving.Also known as Newton's second law of motion.

Force = mass\times Acceleration

F= m × a

As we can see, that force is directly proportional to the mass of the an object.

Force ∝ mass of an object

More the mass of an object more will be the magnitude of the force and vice-versa.

So, from the given objects the mass of the basketball is largest that is 600 g due to which it will take most of the force while throwing it.

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PLEASE HELP!
lana66690 [7]

Answer:

x = 4.32 [m]

Explanation:

We must divide this problem into three parts, in the first part we must use Newton's second law which tells us that the force is equal to the product of mass by acceleration.

∑F = m*a

where:

F = force = 700 [N]

m = mass = 2030 [kg]

a = acceleration [m/s²]

Now replacing:

F=m*a\\700=2030*a\\a = 0.344[m/s^{2}]

Then we can determine the final speed using the principle of conservation of momentum and amount of movement.

(m_{1}*v_{1})+Imp_{1-2}=(m_{1}*v_{2})

where:

m₁ = mass of the car = 2030 [kg]

v₁ = velocity at the initial moment = 0 (the car starts from rest)

Imp₁₋₂ = The impulse or momentum (force by the time)

v₂ = final velocity after the impulse [m/s]

(2030*0) + (700*5)=(2030*v_{2})\\3500 = 2030*v_{2}\\v_{2}=1.72[m/s]

Now using the following equation of kinematics, we can determine the distance traveled.

v_{2}^{2} =v_{1}^{2}+2*a*x

where:

v₂ = final velocity = 1.72 [m/s]

v₁ = initial velocity = 0

a = acceleration = 0.344 [m/s²]

x = distance [m]

1.72^{2}=0^{2} +(2*0.344*x) \\2.97 = 0.688*x\\x = 4.32 [m]

8 0
3 years ago
SOH-CAH-TOA is used to solve for the ________
Misha Larkins [42]

Answer:

c. initial (x and y)

Explanation:

When a projectile is launched at a velocity with a launch angle, to solve it, we must first resolve the initial velocity into the x and y components. To do this will mean we have to treat it like a triangle due to the launch angle and the direction of the projectile.

Therefore, we will have to make use of trigonometric ratios which is also known by the mnemonic "SOH CAH TOA"

Thus, this method resolves the initial x and y velocities.

3 0
3 years ago
If a substance has a pH of 10 -11, it is considered a ...
Mrac [35]

Answer:

Base

Explanation:

anything with a pH between 0-6 is an acid.  pH of 7 means neutral.  8-14 means it is a base

hope I helped :)

8 0
4 years ago
The cannon on a battleship can fire a shell a maximum distance of 36.0 km.
Paraphin [41]

(a)The initial velocity of the shell will be 594.27 m/sec

(b)The maximum height it reaches will be 9000 m.

c)101.249 m meters lower will its surface be 36.0 km from the ship along a horizontal line parallel to the surface of the ship.

d)The error could be significant compared to the size of a target. Option C is correct.

<h3>What is velocity?</h3>

The change of distance with respect to time is defined as speed. Speed is a scalar quantity. It is a time-based component. Its unit is m/sec.

The given data in the problem is

m is the mass of the block = Kg.

u is the initial velocity of fall = m/sec

h is the distance of fall =  m

g is the acceleration of free fall = m/sec²

v is the hitting velocity of =?

a)

The range of the projectile is;

\rm R = \frac{u^2 sin 2 \theta }{g} \\\\ 36 \times 10^ 3 = \frac{U^2 sin 45^0}{9.81} \\\\ U= 594.27  \ m/sec

b)

The maximum height of the projectile is;

\rm H = \frac{u^2 sin 2 \theta }{2g} \\\\ H = \frac{(594.27)^2\times (sin 45)^2}{2 \times 9.81 } \\\\ H = 9000 \ m

c)

The distance between its surface and the ship, measured in a horizontal arc parallel to the surface, will be 36.0 kilometers. The distance from the lower surface is found as;

\rm( R_e + h)^2 = R_e^2+(36)^2 \\\\ (R_e)^2 = h^2+2R_e h= R_e^2 + 12196 \\\\ h^2 + 12800 h - 1296 = 0 \\\\ h = 101.249 \ m

d)

An error is a mistaken or erroneous action. In some contexts, an error is interchangeable with a mistake.

The difference between the calculated value and the original value is known as the error. The inaccuracy may be large in comparison to the target's size. Option C is correct.

Hence the initial velocity of the shell, maximum height, and the distance from the lower surface will be 594.27 m/sec,9000 m, and 101.249 m and option c for question d are correct respectively.

To learn more about the velocity, refer to the link: brainly.com/question/862972.

#SPJ1

4 0
2 years ago
do you think the government should eencourage more aquaculture, the use of new fish species, or both? explain
vodomira [7]
I believe the government should encourage both children(society) do not know much about either because they are not taught enough about that subject/topic.
3 0
3 years ago
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