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Julli [10]
4 years ago
12

Which statement might be correct based on dimensional analysis?

Physics
1 answer:
ss7ja [257]4 years ago
8 0

Answer:

none

Explanation:

a. the unit of height does not carry a square( m^2 is unit of area)

b. time cannot be in m/s. it is unit of speed

c. m/s^2 are the units of acceleration

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Cognitive,emotional, and physical reasons why adolescents engage in riskier behavior
taurus [48]
Cognitive:
This is because teens tend to excrete more hormones when they are in their adolescent ages causing them to act in a riskier behavior.
Emotions:
Because of their feeling that they want to have freedom.
Physical:
Their hormones are affecting their actions.
8 0
3 years ago
A box of mass m1 = 20.0 kg is released from rest at a warehouse loading dock and slides down a 3.0 m-high frictionless chute to
Harrizon [31]

´To develop this problem we will use the concepts related to the conservation of momentum and the application of energy conservation equations to find the velocity of the mass after the collision, like this:

Velocity of the mass m_1 just before the collision

v_1 = \sqrt{2gh}

v_1 = \sqrt{2(9.8)(3)}

v_1 = 7.67m/s

Therefore the momentum just before collision would be

p_2 = m_1v_1+40(0)\\p_1 = 20*7.67+40(0)\\p_1 = 153.36kg \cdot m/s

Momentum after the collision

p_1 = 20*u_1+40u_1\\p_1 = 60u_1

Since the momentum is conserved we have that

153.36= 60u_1

u_1 = \frac{153.36}{60}

u_1 = 2.56m/s

The velocity of mass m_2 after the collision is given by

v_2 = \frac{2m_1}{m_1+m_2} u_1

v_2 = \frac{2(20)}{20+40}(2.56)

v_2 = 1.71m/s

Therefore the change in momentum of mass 2 is

p_2 = m_2v_2

p_2 = 40*1.71

p_2 = 68.4kg\cdot m/s

Therefore the impulse acting on m2 during the collision between the two boxes is p = 68.4kg\cdot m/s

8 0
4 years ago
Find the initial velocity of a ball that has a final velocity of 17.9 m/s, west after it accelerated at 63.7 m/s2, west for 0.21
Musya8 [376]

Explanation:

Taking west to be positive, given:

v = 17.9 m/s

a = 63.7 m/s²

t = 0.21 s

Find: v₀

v = at + v₀

17.9 m/s = (63.7 m/s²) (0.21 s) + v₀

v₀ = 4.52 m/s

The initial velocity is 4.52 m/s west.

5 0
3 years ago
AEROSPACE On the Moon, a falling object falls just 2.65 feet in the first second after being dropped. Each second it falls 5.3 f
aniked [119]

Answer:

d = 265 ft

Therefore, an object fall 265 ft in the first ten seconds after being dropped

Explanation:

This scenario can be represented by an arithmetic progression AP.

nth term = a + nd

Where a is the first term given as 2.63 ft.

d is the common difference and is given as 5.3ft.

n is the particular second/time.

To calculate how far the object would fall in the first 10 seconds, we can derive it using the sum of an AP.

d = nth sum = (n/2)(2a+(n-1)d)

Where n = 10 seconds

a = 2.65 ft

d = 5.3 ft

Substituting the values we have;

d = (10/2)(2×2.65 + (10-1)5.3)

d = 265 ft

Therefore, an object fall 265 ft in the first ten seconds after being dropped

4 0
3 years ago
Read 2 more answers
NEED THE ANSWER ASAP!!
vesna_86 [32]

the earth is on the month of June

8 0
3 years ago
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