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Ronch [10]
3 years ago
9

Sarah throws a ball directly upward at the edge of a cliff with a starting velocity of 3.0 m/s

Physics
1 answer:
strojnjashka [21]3 years ago
6 0

Answer:

The correct answer is t = 0.92s

Explanation:

Initial velocity v​0 = 3.0 m/s

Displacement Δy = ?

Acceleration a = -9.8m/s2

Final velocity v = -6.0m/s

Time t=? Target unknown

We can use the kinematic formula missing Δy to solve for the target unknown t:

V=v0+at

We can rearrange the equation to solve to t:

V-v0=at

t= v-v0/a

Substituting the known value into the kinematic formula gives:

t= (-6.0m/s)-(3.0m/s)

————————————

-9.8m/s2

= -9m/s

—————-

-9.8m/s2

=0.92s

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An 81.5-kg man stands on a horizontal surface.
OLga [1]

Answer:

a)V= 0.0827 m³

b)P=181.11 x 10²  N/m²

Explanation:

Given that

m = 81.5 kg

Density ,ρ = 985 kg/m³

As we know that

Mass = Volume x Density

81.5 = V x 985

V= 0.0827 m³

The force exerted by weight = m g

 F= m g= 81.5 x 10 = 815 N      ( Take ,g= 10 m/s²)

Area ,A= 4.5 x 10⁻² m²

The Pressure P

P=\dfrac{F}{A}

P = \dfrac{815}{4.5\times 10^{-2}}\ N/m^2

P=181.11 x 10²  N/m²

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3 years ago
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oee [108]
I believe the answer is A) Less work in less time. 
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What is the target heart rate for a 24 year old?
andriy [413]

Answer:

196

Explanation:

subtract 24 from 220 to get your answer.

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2 years ago
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Which example possesses mechanical potential energy?. A. a taut guitar string. B. an oscillating pendulum. C. a roller coaster r
Andrej [43]
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If a 20 g cannonball is shot from a 5 kg cannon with a velocity of 100
balu736 [363]

Strange as it may seem, the statement in the question appears to be <em>TRUE</em>.  

-- Before the shot, neither the cannon nor the ball is moving, so their combined momentum is zero.  

-- Since momentum is conserved, we know immediately that their combined momentum AFTER the shot also has to be zero.

-- (20g is rather puny for a "cannonball" ... about the same weight as four nickels. But we'll take your word for it and just do the Math and the Physics.)

-- Momentum = (mass) x (velocity)

After the shot, the momentum of the cannonball is

(0.02 kg) x (100 m/s ==> that way)

Momentum of the ball = 2 kg-m/s ==> that way.

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Momentum of cannon = (5 kg) x (V m/s this way <==)

2 kg-m/s this way <== = (5 kg) x (V m/s this way <==)

Divide each side by (5 kg):

V m/s  =  (2/5) m/s this way <==

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