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nikitadnepr [17]
3 years ago
12

3. A car has a mass of 1,000 kilograms. If a net force of 2,000 N is exerted on the car, what is its acceleration?

Physics
1 answer:
BartSMP [9]3 years ago
7 0

Answer:

2 m/s^2

Explanation:

Acceleration = Force/mass

= 2,000/1,000

= 2

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A rocket sled for testing equipment under large accelerations starts at rest and accelerates according to the expression a = (3.
mash [69]

Answer:

9.6 m

Explanation:

This is a  case of motion under variable acceleration . So no law of motion formula will be applicable here. We shall have to integrate the given equation .

a = 3.6 t + 5.6

d²x / dt² = 3.6 t + 5.6

Integrating on both sides

dx /dt = 3.6 t² / 2 + 5.6 t + c

where c is a constant.

dx /dt = 1.8  t²  + 5.6 t + c

when t = 0 , velocity dx /dt is zero

Putting these values in the equation above

0 = 0 +0 + c

c = 0

dx /dt = 1.8  t²  + 5.6 t

Again integrating on both sides

x = 1.8 t³ / 3 + 5.6 x t² /2 + c₁

x = 0.6 t³  + 2.8  t²  + c₁

when t =0,  x = 0

c₁ = 0

x = 0.6 t³  + 2.8  t²  

when t = 1.6

x = .6 x 1.6³ + 2.8 x 1.6²

= 2.4576 + 7.168

= 9.6256

9.6 m

5 0
3 years ago
When the drivers pass each other, the driver of the red car is to toss a package of contraband to the other driver. To catch the
Brrunno [24]

Answer:

D=387.28m

Explanation:

At the moment where the toss is made X_R = X_G, so we need both equations:

For the red car:

X_R=\frac{a_R*t^2}{2}   With initial speed of 0 and acceleration of 6.12m/s^2.

For the green car:

X_G=Xo + V_G*t   With V_G = 60km/h*\frac{1000m}{1km} * \frac{1h}{3600s} = 16.66m/s   and Xo = 200m

Since both positions will be the same:

\frac{a_R*t^2}{2}=Xo+V_G*t   Solving for t:

t1 = -5.8s  and   t1 =11.25s

Replacing t = 11.25 on either equation to find the displacement:

D = X_R = \frac{a_R*t^2}{2} = 387.28m

3 0
4 years ago
when hard stabilization structures such as groins are used to stabilize a shoreline, the change in the longshore current results
Nikitich [7]

When hard stabilization structures such as groins are used to stabilize a shoreline, the change in the longshore current results <u>deposition of sediment. </u>

On the upcurrent side of the barrier, sediment is deposited as the longshore current slows.

What is Hard stabilization?

  • Hard stabilization is the prevention of erosion through the use of artificial barriers.
  • Other hard stabilization structures, such as breakwaters and seawalls, are built parallel to the beach to protect the coast from the force of waves.
  • Hard stabilization structures, such as groins, are built at right angles to the shore to prevent the movement of sand down the coast and maintain the beach.
  • These constructions are made to last for many years, but because they detract from the visual splendor of the beach, they are not always the ideal answer.
  • Additionally, they affect the habitats and breeding sites of native shoreline species, interfering with the ecosystem's natural processes.

Learn more about the Hard stabilization with the help of the given link:

brainly.com/question/16022736

#SPJ4

4 0
2 years ago
A lamp hangs vertically from a cord in a descending elevator that decelerates at 1.7 m/s2. (a) if the tension in the cord is 63
zubka84 [21]

(a)
The formula is: 
∑ F = Weight + T = mass * acceleration 

as the elevator and lamp are moving downward, I choose downward forces to be positive. 
Weight is pulling down = +(9.8 * mass) 
Tension is pulling up, so T = -63 
Acceleration is upward = -1.7 m/s^2 

(9.8 * mass) + -63 = mass * -1.7 
Add +63 to both sides 
Add (mass * 1.7) to both sides 

(9.8 * mass) + (mass * 1.7) = 63 
11.5 * mass = 63

mass = 63 / 11.5 

Mass = 5.48 kg 


(b)
Since the elevator and lamp are going upward, I choose upward forces to be positive. 
Weight is pulling down = -(9.8 * 5.48) = -53.70 
Acceleration is upward, so acceleration = +1.7 


-53.70 + T = 5.48 * 1.7

T = 53.70 + 9.316 = approx 63 N 

The Tension is still the same - 63 N since the same mass, 5.48 kg, is being accelerated upward at the same rate of 1.7 m/s^2

3 0
3 years ago
1 Calculate the size of the quantum involved in the excitation of (a) an electronic motion of frequency 1.0 × 1015 Hz, (b) a mol
torisob [31]

Answer: a) E= 6.63x10^-19J

E= 3.97×10^2KJ/mol

b) E = 3.31×10^-19J

E= 18.8×10^4 KJ/mol

C) E = 1.32×10^-33J

E= 8.01×10^-10KJ/mol

Explanation:

a) E = h ×f

h= planks constant= 6.626×10^-34

E=(6.626×10^-34)×(1.0×10^15)

E=6.63×10^-19J

1mole =6.02×10^23

E=( 6.63×10^-19)×(6.02×10^23)

E=3.97×10^2KJ/mol

b) E =(6.626×10^-34)/(1.0×10^15)

E=3.13×10^-19J

E= 3.13×10^-19) ×(6.02×10^23)

E= 18.8×10^3KJ/MOL

c) E= (6.626×10^-34) /0.5

E= 1.33×10^-33J

E= (1.33×10^-33) ×(6.02×10^23)

E= 8.01×10^-10KJ/mol

8 0
4 years ago
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