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2 years ago

7

A pole vaulter takes 1.3 seconds to fall from peak height to the landing mat. What is his vertical velocity at contact with the

mat?

1 answer:

julia-pushkina [17]2 years ago

6 0

$12.75 \frac{m}{s}$

Assuming vertical acceleration of $9.81 \frac{m}{s^2}$ , the speed after x seconds of falling is $9.81x \to 9.81(1.3) = 12.75$

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A cube that has a volume of 1.00 m³ contains N distinguishable particles.

grandymaker [24]

Answer:

1 P = 0.5

2 P = 0.3

3 P = 0.01

Explanation:

The probability formula is

$P =V^N$

Where P is the probability V is the volume while N is the number of distinguishing particles

So for N = 1 and $V = 0.500m^3$

$P = (0.500m^3)^1\\$

= 0.5

For N = 1 and $V = 0.300m^3$

$P = (0.300m^3)^1$

= 0.3

For N = 1 and $V = 0.0100m^3$

$P =(0.0100m^3)^1$

= 0.01

6 0

3 years ago

If the resistance of a circuit increases, that means that the current has to

Gnoma [55]

Answer:

If resistance increases current decreases.

Explanation:

Current is <em>inversely proportional</em> to the resistance.
from the relation given below, we can clearly see the relation between current and resistance;

V=IR

I ∝ 1/R

This relation shows that when resistance increases,current decreases.

4 0

3 years ago

Which career professionals are part of the Architecture and Construction career cluster? Check all that apply.

oksian1 [2.3K]

Answer:

A,B,D,E,F

Explanation:

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3 0

3 years ago

A 5 kg ball moving to the right at a speed of 6 m/s strikes another 4 kg

Dahasolnce [82]

Answer:

The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.

Explanation:

Momentum before = momentum after

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(5 kg) (6 m/s) + (4 kg) (-5 m/s) = (5 kg) v₁ + (4 kg) v₂

10 m/s = 5 v₁ + 4 v₂

Assuming an elastic collision, kinetic energy is conserved.

½ m₁ u₁² + ½ m₂ u₂² = ½ m₁ v₁² + ½ m₂ v₂²

m₁ u₁² + m₂ u₂² = m₁ v₁² + m₂ v₂²

(5 kg) (6 m/s)² + (4 kg) (-5 m/s)² = (5 kg) v₁² + (4 kg) v₂²

280 m²/s² = 5 v₁² + 4 v₂²

Substituting:

v₂ = (10 − 5 v₁) / 4

280 = 5 v₁² + 4 [(10 − 5 v₁) / 4]²

280 = 5 v₁² + (10 − 5 v₁)² / 4

1120 = 20 v₁² + (10 − 5 v₁)²

1120 = 20 v₁² + 100 − 100 v₁ + 25 v₁²

0 = 45 v₁² − 100 v₁ − 1020

0 = 9 v₁² − 20 v₁ − 204

0 = (9 v₁ + 34) (v₁ − 6)

v₁ = -3.78 m/s or 6 m/s

u₁ = 6 m/s, so v₁ = -3.78 m/s. Solving for v₂:

v₂ = (10 − 5 v₁) / 4

v₂ = 7.22 m/s

The 5 kg ball moves 3.78 m/s to the left, and the 4 kg ball moves 7.22 m/s to the right.

6 0

3 years ago

The net external force on a golf cart is 411 N north of tha cart has a total mass of 281 kg what is tha cart acceleration

Alisiya [41]

1.46m/s²

Explanation:

Given parameters:

Mass of cart = 281kg

Net external force = 411N

Unknown:

Acceleration of the cart = ?

Solution:

According to newton's second law " the net force on a car is the product of its mass and acceleration";

Force = mass x acceleration

Since acceleration of the cart is unknown;

Acceleration = $\frac{force}{mass}$ = $\frac{411}{281}$

Acceleration = 1.46m/s²

learn more:

Newton's laws brainly.com/question/11411375

#learnwithBrainly

8 0

3 years ago