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Aleksandr-060686 [28]
3 years ago
12

A carmaker has designed a car that can reach a maximum acceleration of 12 meters/second2. The car’s mass is 1,515 kilograms. Ass

uming the same engine is used, what should the car’s mass be if the carmaker wants to reach an acceleration of 15 meters/second2? Use F = ma.
A.
1,212 kg
B.
1,335 kg
C.
1,466 kg
D.
1,515 kg
E.
1,894 kg
Physics
2 answers:
Jet001 [13]3 years ago
8 0

Answer:

A: 1,212 kg

Explanation:

irga5000 [103]3 years ago
3 0

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kasasasajjas

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The weakest gravitational force caused by the Moon is called the __________, which is found at the point on Earth farthest from
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This is called Nadir.
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2 years ago
A dog barking at the sound of the postal carrier delivering mail is reacting to what type of cue?
Gwar [14]

Answer:

A. External

Explanation:

External stimulus includes touch/pain, vision, smell, taste, and sound.

6 0
2 years ago
Despite a very strong wind, a tennis player manages to hit a tennis ball with her racquet so that the ball passes over the net a
kobusy [5.1K]

Answer:

The downward force of gravity and the force exerted by the air.

Explanation:

Gravity, in a Newtonian Framework, acts at distance, so, the ball doesn't has to be connected to Earth to feel its gravitational pull (as we know by experience).

The racket only can exert a force in the ball when is touching it, so, once the ball has left the contact with the racket, there is no force by the "hit".

The air is always touching the ball on the atmosphere (even when physicist pretend that is not). So, there is drag from the wind, and a buoyant force  exerted by the air over the ball at every time.

5 0
4 years ago
The count rate of a radioactive source decreases from 1600 counts per minute to 400 counts per minute in 12 hours. What is the h
kirill115 [55]

Answer:

t_{1/2}=6 h

Explanation:

Let's use the decay equation.

A=A_{0}e^{-\lambda t}

Where:

  • A is the activity at t time
  • A₀ is the initial activity
  • λ is the decay constant

We know that \lambda=\frac{ln(2)}{t_{1/2}}

So we have:

\lambda=\frac{ln(A/A_{0})}{t}

\frac{ln(2)}{t_{1/2}}=\frac{ln(A/A_{0})}{t}

t_{1/2}=\frac{t*ln(2)}{ln(A/A_{0})}

t_{1/2}=6 h

Therefore, the half-life of the source is 6 hours.

I hope it helps you!

4 0
3 years ago
A projectile is fired horizontally from a height of 10 m above level ground. The projectile lands a horizontal distance of 15 m
Degger [83]

Answer:

a)t  = 1,43 s

b) V = 10,49 m/s

c) V₀ₓ = 10,49 m/s   ;    V₀y = 14,01 m/s

d) Vf = 17,5 m/s

Explanation:

According to the problem statement

V₀ = V₀ₓ    and  V₀y = 0

And at the end of the movement t = ?  the distance y = 10 m

Therefore as

h = V₀y - (1/2)*g*t²

Vertical distance y = h = 10 = V₀y*t - 0,5 (-9,8)*t²

10 = 4,9*t²

t² = 10/4,9    ⇒  t² = 2,04 s

t  = 1,43 s

a) 1,43 s is the time of movement

b) V₀ = V₀ₓ        V₀y = 0     and  V₀ₓ = Vₓ     ( constant )

Just before touching the ground, the horizontal distance is

hd = 15 = Vₓ * t

Then  15 /1,43 = Vₓ = V₀ₓ

Vₓ = 10,49 m/s

Then initial speed is V = 10,49 m/s    since V₀y = 0

Vf² = Vₓ² + Vy²

Vyf = V₀y - g*t

Vyf =  0 - 9,8 *1,43

Vyf = - 14,01 m/s

And finally the speed when the projectile strike the ground is:

Vf² = Vₓ² + Vy²

Vf = √ (10,49)² + (14,01)²

Vf = 17,50 m/s

3 0
3 years ago
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