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Veseljchak [2.6K]
3 years ago
11

ASAP NEED HELP PLEASE

Physics
1 answer:
Sholpan [36]3 years ago
5 0

Answer:

57.91

Explanation:

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aleksandrvk [35]
That statement is wrong.

Domestic violence could happen to anyone.

There are a lot of cases in which the domestic violence's victims are men or elderly people.
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3 years ago
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A man of weight Wman is standing on the second floor and is pulling on a rope to lift a box of weight Wbox from the floor below.
slavikrds [6]

Answer:

Explanation:

See the attached figure . See the forces acting on man pulling up the box .

Man is stationary so net force acting on man is zero .

T + R = Wman

R is the reaction force of the ground of second floor  .

R = Wman - T

3 0
3 years ago
Solve the inequality. x/3 is greater than or equal to - 6. a. x ≥ –9 b. x ≥ 9 c. x ≥ –18 d. x ≤ –18
NISA [10]
To get x on its own, you times the 3 over to the other side so the 3 cancels out on the LHS. 

~ x greater than or equal to -18

(C)
6 0
3 years ago
An 80- quarterback jumps straight up in the air right before throwing a 0.43- football horizontally at 15 . How fast will he be
lord [1]

Answer:

a)

the quarterback will be moving back at speed of 0.080625 m/s

b)

the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

Explanation:

Given the data in the question;

a)

How fast will he be moving backward just after releasing the ball?

using conservation of momentum;

m₁v₁ = m₂v₂

v₂ = m₁v₁ / m₂

where m₁ is initial mass ( 0.43 kg )

m₂ is the final mass ( 80 kg )

v₁ is the initial velocity  ( 15 m/s )

v₂ is the final velocity

so we substitute

v₂ = ( 0.43 × 15 ) / 80

v₂ = 6.45 / 80

v₂ = 0.080625 m/s

Therefore, the quarterback will be moving back at speed of 0.080625 m/s

b) Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?

we make use of the relation between time, distance and speed;

s = d/t

d = st

where s is the speed ( 0.080625 m/s )

t is time ( 0.30 s )

so we substitute

d = 0.080625 × 0.30

d = 0.0241875 m or 2.41875 cm

Therefore, the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm

5 0
3 years ago
A certain first-order reaction is 58% complete in 95 s. What are the values of the rate constant and the half-life for this proc
guajiro [1.7K]

Answer:

0.005734 s^{-1} and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..

Explanation:

Expression for rate law for first order kinetics is given by:

a_o=a\times e^{-kt}

where,

k = rate constant  

t = age of sample

a_o = let initial amount of the reactant  

a = amount left after decay process  

We have :

a_o=x

a=58\%\times x=0.58x

t = 95 s

0.58x=x\times e^{-k\times 95 s}

\k= 0.005734 s^{-1}

Half life is given by for first order kinetics::

t_{1/2}=\frac{0.693}{k}

=\frac{0.693}{0.005734 s^{-1}}=120.86 s

0.005734 s^{-1} and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..

3 0
3 years ago
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