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leva [86]
3 years ago
6

He likes to eat (into passive voice)​

Physics
1 answer:
Mama L [17]3 years ago
6 0

Answer:

yes he does

Explanation:

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What is the main function of b-cells?
RoseWind [281]

Answer: They create antibodies.

7 0
3 years ago
A rocket initially at rest accelerates at a rate of 99.0 meters/second ^2.Calculate the distance covered by the rocket if it. At
Pani-rosa [81]

The distance covered is 1000 m

Explanation:

The rocket is moving by uniformly accelerated motion, so we can find the distance it covers by using the following suvat equation:

s=vt-\frac{1}{2}at^2

where

s is the distance covered

v is the final velocity

t is the time

a is the acceleration

For the rocket in this problem, we have:

v = 445 m/s is the final velocity

a=99.0 m/s^2 is the acceleration

t = 4.50 s is the time

Substituting, we find the distance covered:

s=(445)(4.50)-\frac{1}{2}(99.0)(4.50)^2=1000 m

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

8 0
4 years ago
A car is traveling at 50 mi/h when the brakes are fully applied, producing a constant deceleration of 44 ft/s2. What is the dist
elena55 [62]

Answer:

Distance, d = 61.13 ft

Explanation:

It is given that,

Initial speed of the car, u = 50 mi/h = 73.34 ft/s

Finally, it stops i.e. v = 0

Deceleration of the car, a=-44\ ft/s^2

We need to find the distance covered before the car comes to a stop. Let the distance is s. It can be calculated using third law of motion as :

v^2-u^2=2as

s=\dfrac{v^2-u^2}{2a}

s=\dfrac{0-(73.34\ ft/s)^2}{2\times -44\ ft/s^2}

s = 61.13 ft

So, the distance covered by the car before it comes to rest is 61.13 ft. Hence, this is the required solution.

4 0
4 years ago
PLEASE HELP I GIVE BRAINLIEST AND 15 POINTS
deff fn [24]
The mass of the man would remain the same…
His weight would change. Assuming the man’s weight on earth is 60N.
Since weight is a force, according to Newton’s second law, F = ma ( m = mass, a = acceleration due to gravity) First lets find the mass of the man, as it is required to find his weight on the moon.
F=ma[taking a of earth as10m/s
2
]
60=m.10[divide10onbothsides]
m=
10
60
​
= 6 Kg
Acceleration due to gravity on the moon is83%less than the acceleration due to gravity on earth(1.622m/s
2
).
F=ma
F=6.1.622=9.732 N
So a person weighting 60N on earth would approximately weight around 10Non the moon.
8 0
3 years ago
A force of 150 N is applied to a 25 cm2 piston in a hydraulic machine. What is the area of a larger piston if the force availabl
Colt1911 [192]

Answer:

The answer to your question is 75 cm²

Explanation:

To solve this problem, we need to use the Pascal's principle that states that the pressure in a point is equal to the pressure in a second point.

                  \frac{F1}{A1} = \frac{F2}{A1}

Process

1.- Solve for A2

                  A2 = \frac{F2A1}{F1}

2.- Substitution

                  A2 = \frac{450N x 25 cm2}{150 N}

3.- Simplification and result

                  A2 = \frac{11250}{150}

                         A2 = 75 cm²

8 0
4 years ago
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