We dont see objects. We see the light ___

The system below has a friction force of 25 N acting on the cart which 8 kg. The mass hanging off the edge has a mass of 6 kg. F

photoshop1234 [79]

The cart will be pulled to the right by the hanging mass, so by Newton's second law, the net force on the cart is

T - 25 N = (8 kg) a

where T is the tension in the rope and a is the acceleration.

The hanging mass has a net force of

(6 kg) g - T = (6 kg) a

where g = 9.8 m/s².

Adding these equations together eliminates T, and we can solve for a :

(T - 25 N) + ((6 kg) g - T ) = (14 kg) a

33.8 N = (14 kg) a

a = (33.8 N) / (14 kg) ≈ 2.4 m/s²

Then the tension in the rope is

T - 25 N = (8 kg) (2.4 m/s²)

T ≈ 25 N + 19.31 N ≈ 44 N

5 0

3 years ago

1) A 1,600 kilogram car is also traveling in a straight line. Its momentum is 32,500 kg*m/s. What is the

BaLLatris [955]

Answer:

v = 20.31 m/s

Explanation:

p = mv -> v = p/m = 32,500 kg*m/s / 1,600 kg = 20.31 m/s

4 0

3 years ago

Vertical Motion

harina [27]

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3 0

3 years ago

If a car go from 0 to 60 mi/hr in 8.0 seconds, what would be its final speed after 5.0 seconds if its starting speed were 50 mi/

Taya2010 [7]

Answer:

v = 87.57 m/s

Explanation:

Given,

The initial velocity of the car, u = 0

The final velocity of the car, v = 60 mi/hr

The time period of car, t = 8 s

= 0.00222 hr

The acceleration of the car is given by the formula,

a = (v -u) / t

= 60 / 0.00222

= 27027 mi/hr²

If the car has initial velocity, u = 50 mi/hr

The time period of the car, t = 5.0 s

= 0.00139 hr

Using first equations of motion

v = u + at

= 50 + (0.00139 x 27027)

= 87.57 mi/hr

Hence, the final velocity of the car, v = 87.57 mi/hr

3 0

3 years ago

A 50.1 kg diver steps off a diving board and drops straight down into the water. The water provides an average net force of resi

Varvara68 [4.7K]

Answer:

The total distance is 16.9 m

Explanation:

We understand work in physics as certain force exerted through certain distance. To reach that point below the water, the work done by the diver must be equal to the work done by the water's force of resistance. Therefore, we determine both work expressions and we solve the equation for the diver distance, which is the total distance between the diving board and the stopping point underwater.

$W_{d}: work done by diver\\W_{R}: work due the force of resistance\\W_{d} = W_{R}\\F_{d}\times d_{d}=F_R \times d_{R}\\d_{d}= \frac{F_R \times d_R}{F_d}= \frac{F_R \times d_R}{m_d \times g}= \frac {1598 N \times 5.2 m}{50.1 kg \times 9.81 m/s^2}\\d_{d}= 16. 91 m$

7 0

3 years ago

We dont see objects. We see the light ____ off objects.