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dem82 [27]
3 years ago
6

5. A box weighs 196 N. A rope is tied to the box. What is the

Physics
1 answer:
natta225 [31]3 years ago
5 0

Answer:

296 N

Explanation:

Draw a free body diagram.  The box has two forces on it: tension up and weight down.

Apply Newton's second law:

∑F = ma

T − mg = ma

T = m (g + a)

Given m = 196 N / 9.8 m/s² = 20 kg, and a = +5 m/s²:

T = (20 kg) (9.8 m/s² + 5 m/s²)

T = 296 N

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A cup sits on a table. Due to its position, the potential energy of the cup is 3.00 joules. Ignoring frictional effects, if the
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Due to conservation of energy, half way the potential energy will be 1.5J so the remaining 1.5J is kinetic energy.
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4 years ago
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Using the formula 1/2 (m x v2), what is the kinetic energy of a 4 kg rock falling through the air at 5 m/s
aivan3 [116]

Answer:

KE = 50J

Explanation:

KE = \frac{1}{2}mv^{2} \\\\ KE = \frac{1}{2}(4)(5)^{2} \\\\ KE = 2(25) \\\\ KE = 50J

5 0
3 years ago
A. A child is twirling a 1.52 kg object in a vertical circle with a radius of 67.6
steposvetlana [31]

Answer:

(a) 4.21 m/s

(b) 24.9 N

Explanation:

(a) Draw a free body diagram of the object when it is at the bottom of the circle.  There are two forces on the object: tension force T pulling up and weight force mg pulling down.

Sum the forces in the radial (+y) direction:

∑F = ma

T − mg = m v² / r

v = √(r (T − mg) / m)

v = √(0.676 m (54.7 N − 1.52 kg × 9.8 m/s²) / 1.52 kg)

v = 4.21 m/s

(b) Draw a free body diagram of the object when it is at the top of the circle.  There are two forces on the object: tension force T pulling down and weight force mg pulling down.

Sum the forces in the radial (-y) direction:

∑F = ma

T + mg = m v² / r

T = m v² / r − mg

T = (1.52 kg) (4.21 m/s)² / (0.676 m) −  (1.52 kg) (9.8 m/s²)

T = 24.9 N

6 0
3 years ago
A clothes dryer uses about 5 amps of current from a 240 volt line. How much power does it use?
yan [13]
PVI, so the power used would be 240*5= 1200 Watts.
6 0
3 years ago
A 500 g model rocket is on a cart that is rolling to the right at a speed of 3.0 m/s . The rocket engine, when it is fired, exer
ivanzaharov [21]

Answer:

The rocket has to be launched 8 m from the hoop

Explanation:

Let's analyze this problem, the rocket is on a car that moves horizontally, so the rocket also has the same speed as the car; The initial horizontal rocket speed is (v₀ₓ = 3.0 m/s).

On the other hand, when starting the engines we have a vertical force, which creates an acceleration in the vertical axis, let's use Newton's second law to find this vertical acceleration

    F -W = m a

    a = (F-mg) / m

    a = F/m  -g

    a = 7.0/0.500  - 9.8

    a = 4.2 m/s²

We see that we have a positive acceleration and that is what we are going to use in the parabolic motion equations

Let's look for the time it takes for the rocket to reach the height (y = 15m) of the hoop, when the rocket fires its initial vertical velocity is zero (I'm going = 0)

    y = v_{oy} t + ½ a t²

    y = 0 + ½ a t²

    t = √ 2y/a

    t = √( 2 15 / 4.2)

    t = 2.67 s

This time is also the one that takes in the horizontal movement, let's calculate how far it travels

    x = v₀ₓ t

    x = 3 2.67

    x = 8 m

The rocket has to be launched 8 m from the hoop

8 0
3 years ago
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