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insens350 [35]
4 years ago
13

True or False: The absorption of alcohol can be slowed down by eating, but only water can reduce the BAL level.

Physics
2 answers:
Andre45 [30]4 years ago
5 0

Answer:

False

Explanation:

The blood alcohol level cannot be changed by anything but the metabolism or the individual that is drinking the alcohol, and that is a job performed by the liver, so no one can speed up that process, alcohol absorption does change depending on what the person has in their stomach, that is why people get dunker mmore easily when they haven´t have anything to eat before drinking.

lesya [120]4 years ago
3 0
A hopes this helps idk this
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During a braking test, a car is brought to rest beginning from an initial speed of 60 mi/hr in a distance of 120 ft. With the sa
maw [93]

Answer:

Explanation:

Given

Initial speed u=60\ mi/hr\approx 88\ ft/s

distance traveled before coming to rest d_1=120\ ft

using equation of motion

v^2-u^2=2as

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

0-(88)^2=2\times a\times 120---1

for u_2=80\ mi/hr\approx 117.33\ ft/s

using same relation we get

0-(117.33)^2=2\times a\times (d_2)----2

divide 1 and 2 we get

(\frac{88}{117.33})^2=\frac{120}{d_2}

d_2=213.32\ ft

So a distance if 213.32 ft is required to stop the vehicle with 80 mph speed

8 0
3 years ago
The system below uses massless pulleys and ropes. The coefficient of friction is μ. Assume that M1 and M2 are sliding. Gravity i
Archy [21]

Explanation:

Using Newtons second law on each block

F = m*a

Block 1

T_{1} - u*g*M_{1} = M_{1} *a \\\\T_{1} = M_{1}*(a + u*g) ... Eq1

Block 2

T_{2} - u*g*M_{2} = M_{2} *a \\\\T_{2} = M_{2}*(a + u*g) ... Eq2

Block 3

- (T_{1} + T_{2} ) + g*M_{3} = M_{3} *a \\\\T_{1} + T_{2} = M_{3}*( -a + g) ... Eq3

Solving Eq1,2,3 simultaneously

Divide 1 and 2

\frac{T_{1} }{T_{2}} = \frac{M_{1}*(a+u*g)}{M_{2}*(a+u*g)}  \\\\\frac{T_{1} }{T_{2}} = \frac{M_{1} }{M_{2} }\\\\ T_{1} =  \frac{M_{1} *T_{2} }{M_{2} } .... Eq4

Put Eq 4 into Eq3

T_{2} = \frac{M_{3}*(g-a) }{1+\frac{M_{1} }{M_{2} } }  ...Eq5

Put Eq 5 into Eq2 and solve for a

a = \frac{M_{3}*g -u*g*(M_{1} + M_{2}) }{M_{1} + M_{2} + M_{3} }  .... Eq6

Substitute back in Eq2 and use Eq4 and solve for T2 & T1

T_{2} = M_{2}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\T_{1} = M_{1}*M_{3}*g*(\frac{1-u}{M_{1} + M_{2}+M_{3}})\\\\

5 0
4 years ago
A muscle builder holds the ends of a masslessrope. At the
Grace [21]

Answer:

Balancing of forces,

In the X-direction:

-Tcos4.5^o +Tcos4.5^o=0

In the Y-direction:

Tsin4.5^o +Tsin4.5^o-m*g=0

2Tsin4.5^o=15*9.81

T=(15*9.81)/(2sin4.5^o)

=937.75 N

Therefore, tension in the rope is 937.75 N.

7 0
3 years ago
QUESTION 1
frez [133]
Go90 day would work better if they could just pay ate a little bit more than we
4 0
3 years ago
In a position vs. time graph, which of the following relationships is true? O The steeper the line the slower the velocity O The
Blizzard [7]

Answer:

dd

Explanation:

5 0
3 years ago
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