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viva [34]
1 year ago
15

A car is traveling as its speed changes from rest to 25 m/s in 5 s. What is the acceleration of the race car in m/s2?

Physics
2 answers:
Leona [35]1 year ago
5 0

Answer:

a=2 m/s²

Explanation:

v=u+at

25=15 +5a

5a=10

a=2 m/s²

Pachacha [2.7K]1 year ago
4 0
The answer is a=2 m/s^2
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A projectile is fired from the ground (you can assume the initial height is the same as the ground) in a field so there are no o
I am Lyosha [343]

Answer:

(A)  111.77m

(B) 9.07m

(C)55.88m/s

(D)20.54m/s

Explanation:

step 1 " we have to calculate the time it took the projectile to get to its maximum height

(a)  t = usinθ/g

= 49sin 33/9.81

49×0.5446/9.81

=2.72s

the horizontal distance =  ucosθ×t  , because the projectile horizontal motion is unaffected by the force of graavity

= 49cos33 ×2.72

=111.77m

(B) with the same projectile fired the same way , the horizontal distance = 55.8m

55.8 = ucosθ×t

55.8 = 49cos33 ×t

t = 55.8/49cos33

t= 1.36s

height of the projectile = 1/2 ×g×t²

=1/2 ×9.81×1.36²

= 9.07m

(c) Velocity of the projectile when it hits the wall

V₀ = ucosθ×t

49cos 33 × 1.36

=55.88 m/s

(D) speed = distance / time

distance /2×t   ; total time of flight

= 55.88/ 2.72

=20.54m/s

4 0
3 years ago
A block of metal of density 3000 kg/m3 is 2m high and stands on a square base of side 0.5 m. Calculate:
labwork [276]

Answer:

1)  A = 0.25 m², 2) V = 0.5 m³, 3)   m = 1500 kg, 4) W = 14700 N,

5)  P = 58800 Pa

Explanation:

1) The area of ​​the base is square

          A = L²

         A = 0.5²

         A = 0.25 m²

2) The block is a parallelepiped

         V = A h

         V = 0.25 2

          V = 0.5 m³

3) Density is defined

           rho = m / V

           m = rho V

           m = 3000 0.5

           m = 1500 kg

4) The weight of a body is

           W = mg

            W = 1500 9.8

            W = 14700 N

5) The pressure is

             P = F / A

in this case the force is equal to the weight of the body

              P = 14700 / 0.25

              P = 58800 Pa

8 0
3 years ago
If a 2000 kg car is at rest and a 500 N force is applied to it. What acceleration will it experience?
Komok [63]
Net Force = mass * acceleration.
Therefore, Acceleration = Net force / mass
A=2000/500
A=4 m/s^2
4 0
4 years ago
Please help on this one
Murljashka [212]
Just find the area of the graft

4m/s x 5s
=20m

7 0
3 years ago
Read 2 more answers
AP Physics I, shouldn't be too hard.
Nana76 [90]

Answer:

The correct option is;

D. The kinetic energy decreases by 3·m₀·v₀²

Explanation:

The given parameters are;

The mass of object X = m₀

The initial velocity of object X = v₀

The mass of object Y = 2·m₀

The initial velocity of object Y = -2·v₀

By conservation of linear momentum, we have;

The total initial momentum = The total final momentum

Therefore, we have;

The total initial momentum = m₀·v₀ - 2·m₀·2·v₀ = The total final momentum

∴ The total final momentum = -3·m₀·v₀

The total mass of the two object after sticking together = 2·m₀ + m₀ = 3·m₀

Therefore, the velocity of the two objects after collision = (The total final momentum)/(Total mass) = -3·m₀·v₀/(3·m₀) = -v₀

The kinetic energy = 1/2 × Mass × (Velocity)²

Therefore, the kinetic energy after collision = 1/2 × (3·m₀) × v₀² = 3·m₀·v₀²/2

The kinetic energy before collision = 1/2 × m₀ × v₀² + 1/2 × (2·m₀) × (2·v₀)² = (1/2 + 4) × (m₀·v₀²)

∴ The kinetic energy before collision =  9·(m₀·v₀²)/2

The change in kinetic energy = The kinetic energy after collision - The kinetic energy before collision = 3·m₀·v₀²/2 - 9·(m₀·v₀²)/2 = -3·m₀·v₀²

Therefore, the kinetic energy decreases by 3·m₀·v₀².

5 0
3 years ago
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