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lapo4ka [179]
3 years ago
5

An object is moving with an initial velocity of 3.3m/s it is subject to a constant acceleration of 3.7 m/s2 for 10 s. How far wi

ll it have traveled during the time of its acceleration
Physics
1 answer:
egoroff_w [7]3 years ago
8 0

Answer:

218m

Explanation:

Given parameters:

Initial velocity  = 3.3m/s

acceleration  = 3.7m/s²

time   = 10s

Unknown:

How far will it travel during the time of acceleration  = ?

Solution:

We use of the kinematics equations to solve this problem;

          S  = ut  +  \frac{1}{2} at²  

S is the distance

u is the initial velocity

t is the time

a is the acceleration

   So;

            S  = (3.3x10)   +   (\frac{1}{2}  x 3.7 x 10²)  = 218m

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Neon gas is an example of
Murljashka [212]
Neon is an element, and on top of that  a noble gas, which makes it resistant to forming compounds with other elements. The answer is A.
6 0
4 years ago
If carbon has an atomic number of 6, how many protons and neutrons are found in the carbon-14 atom? A.
Alex Ar [27]
The correct answer is B. 6 protons and 8 neutrons

Carbon-14 has same atomic number of 6. It has a nucleon number of 14
Atomic number = proton number = 6
Neutron number = nucleon number - atomic number = 14 - 6 = 8

Hope it helped!
5 0
4 years ago
A ball is dropped from a boat so that it strikes the surface of a lake with a speed of 16.5 ft/s. While in the water, the ball e
denis23 [38]
The initial velocity is
v(0) = 16.5 ft/s

While in the water, the acceleration is
a(t) = 10 - 0.\frac{dv}{dt} =10-0.8v \\\\  \frac{dv}{10-0.8v}=dt \\\\ \int_{16.5}^{v} \,  \frac{dv}{10-0.8v}  = \int_{0}^{t} dt \\\\ - \frac{1}{0.8} [ln(10-0.8v)]_{16.5}^{v}=t \\\\ ln \frac{10-0.8v}{-3.2}=-0.8t \\\\  \frac{0.8v -10}{3.2}  =e^{-0.8t} \\\\ 0.8v = 10 + 3.2e^{-0.8t} \\\\ v=12.5+4e^{-0.08t}

The velocity function is
v(t)=12.5+4e^{-0.8t}
It satisfies the condition that v(0) = 16.5 ft/s.
When t = 5.7s, obtain
v(5.7)=12.5+4e^{-0.8\times5.7} = 12.54 \, ft/s

The depth of the lake is
d=\int_{0}^{5.7} \, (12.5+4e^{-0.8t})dt \\\\ = 12.5(5.7)+ \frac{4}{(-0.8)}[e^{-0.8t}]_{0}^{5.7} \\\\ =71.25-5(0.0105-1) =76.198 \, ft

Answer:
The velocity at the bottom of the lake is 12.5 ft/s
The depth of the lake is 76.2 ft


7 0
3 years ago
If Susan exerted a force of 5 newtons and the output force of the machine is 15
Anon25 [30]

Answer:

Mechanical advantage = 3

Explanation:

Mechanical advantage, MA = Output force/Input force

Output force = 15 N. Input force = 5 N

MA = Output force/Input force = 15 N/ 5 N = 3

4 0
4 years ago
You are riding in an elevator that is accelerating upward. Suppose you stand on a scale. The reading on the scale is __________.
olchik [2.2K]

Answer:

greater than your true weight

Explanation:

When going up in an elevator the acceleration of the elevator is added to the acceleration due to gravity. This will increase the reading on the scale.

The expression of the resultant weight will be

N=m(a+g)

where,

m = Mass of the person

g =  Acceleration due to gravity = 9.81 m/s²

a = Acceleration of the elevator.

Hence, the reading on the scale is <u>greater than your true weight.</u>

5 0
3 years ago
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