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amm1812
4 years ago
8

A long. 1.0 kg rope hangs from a support that breaks, causing the rope to fall, if the pull exceeds 43 N. A student team has bui

lt a 2.0 kg robot "mouse" that runs up and down the rope. What minimum magnitude of the acceleration should the robot have for the rope to fail? Express your answer with the appropriate units.
Physics
1 answer:
raketka [301]4 years ago
8 0

Answer:

6.8 m/s2

Explanation:

Let g = 9.8 m/s2. The total weight of both the rope and the mouse-robot is

W = Mg + mg = 1*9.8 + 2*9.8 = 29.4 N

For the rope to fails, the robot must act a force on the rope with an additional magnitude of 43 - 29.4 = 13.6 N. This force is generated by the robot itself when it's pulling itself up at an acceleration of

a = F/m = 13.6 / 2 = 6.8 m/s2

So the minimum magnitude of the acceleration would be 6.8 m/s2 for the rope to fail

You might be interested in
Select the correct answer.
Nastasia [14]

The force of a test charge would be doubled if the electric field is doubled.

Answer: Option A

<u>Explanation:</u>

Electric field is the region or range up to which a charge particle will have its influence of electric energy on another charged particles. So the experienced force by the test charge up to a certain range is defined as the electric field of that charged particle.

This means that the electric field strength is inversely proportionate to the test charge and directly proportionate to the force acting on the test charge.  As ,

               E=\frac{\text { Force }}{\text { Test charge }}

So, force will be product of electric field strength with test charge.  Thus,

              \text { Force } \propto \text { Test charge and Force } \propto \text { Electric field }

So, if there is increase in the electric field, then there will be increase in the force of the test charge. Thus, if the electric field is doubled thereby the force of a test charge will also be doubled.

4 0
4 years ago
How much voltage is required to run 0.42 A of current through a 150
igomit [66]

The voltage in the resistor is 63 V

Explanation:

We can solve the problem by applying Ohm's law, which states the relationship between voltage, current and resistance in a resistor:

V=RI

where

V is the voltage

R is the resistance

I is the current

For the resistor in this problem, we have:

I = 0.42 A is the current

R=150 \Omega is the resistance

Substituting into the equation, we find the voltage needed:

V=(0.42)(150)=63 V

Learn more about voltage and current:

brainly.com/question/4438943

brainly.com/question/10597501

brainly.com/question/12246020

#LearnwithBrainly

4 0
3 years ago
Read 2 more answers
If you are driving 90 km/hkm/h along a straight road and you look to the side for 2.2 ss , how far do you travel during this ina
ICE Princess25 [194]

Answer: 55m

Explanation:

Given the following :

Driving speed = 90km/hr

Inattentive period (time) = 2.2s

Distance during inattentive period =

(driving speed * time)

Converting driving speed from km/hr to m/s

1000m = 1km

3600s = 1hour

Therefore,

90km/hr = (90 * 1000) / 3600

90km/hr = (90000)/ 3600 = 25m/s

Therefore ;

Distance during inattentive period =

(driving speed * time)

Distance during inattentive period = (25m/s × 2.2s) = 55m

Distance traveled during inattentive period is 55m

6 0
3 years ago
A new electric car makes use of storage batteries as it source of energy. Its mass is 1300kg and it is powered 26 batteries, eac
Lorico [155]

Answer:

a) Horse power required = 3.575 hp

b) The battery must be charged after approximately 225 km

Explanation:

a) Frictional force, F = 240 N

Average speed of the car = 40 km/h = 40 * (1000/3600) = 11.11 m/s

Horse power required = Power dissipated by the frictional force

Power = Force * Velocity

Power = 240 * 11.11

Power = 2666.67 Watt

746 Watt = 1 hp

2666.67 Watt = 2666.67/746

Horse power required = 3.575 hp

b)

The charge of 1 battery, Q = 52 Ah = 52 * 3600 = 187,200 C

Q = 187,200 C

For 24 batteries, Q = 24 * 187,200

Q = 4492800 C

Power consumed by the batteries, P = QV/t

P = 2666.67 W

2666.67 = (4492800 * 12)/t

t =  (4492800 * 12)/2666.67

t = 20217.575 s

Velocity = distance/time

11.11 = distance/20217.575

distance = 20217.575 * 11.11

distance = 224617.26 m

distance = 224.62 km

5 0
3 years ago
An 880 kg cannon at rest fires a 12.4 kg cannonball forward at 540 m/s. What is the recoil velocity of the cannon
Marina86 [1]

Answer:

<em>The recoil velocity of the cannon is 7.61 m/s in the opposite direction of the cannonball</em>

Explanation:

<u>Linear Momentum </u>

The principle of conservation of the linear momentum establishes that the sum of the linear momentums of every object in an isolated system (no external forces) is constant, regardless of the interactions between them.

Let's think we have two objects with masses m_1 and m_2, moving at speeds v_1 and v_2. If they collide and change their speeds to v_1' and v_2', then

m_1v_1+m_2v_2=m_1v_1'+m_2v_2'

In our problem, the 880 kg cannon is initially at rest and has the cannonball of 12.4 Kg inside of it. As the initial speed of both joined objects is zero, the initial total momentum is zero. After the ball is fired, the ball moves at v_2=540 m/s. We need to find the recoil velocity of the cannon v_1'

m_1v_1'+m_2v_2'=0

\displaystyle m_1v_1'=-m_2v_2'

\displaystyle v_1'=-\frac{m_2v_2'}{m_1}

\displaystyle v_1'=-\frac{12.4(540)}{880}=-7.61\ m/s

The recoil velocity of the cannon is 7.61 m/s in the opposite direction of the cannonball

5 0
3 years ago
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