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3 years ago

An unbalanced force of 500 N is applied to a 75 kg object. What is the acceleration of the object?

Physics

2 answers:

Lubov Fominskaja [6]3 years ago

8 0

Answer:

Acceleration, $a=6.66\ m/s^2$

Explanation:

Given that,

Force acting on the object, F = 500 N

Mass of the object, m = 75 kg

We need to find the acceleration of the object. The force acting on the object is given by :

$a=\dfrac{F}{m}$

$a=\dfrac{500\ N}{75\ kg}$

$a=6.66\ m/s^2$

So, the acceleration of the object is $6.66\ m/s^2$ . Hence, this is the required solution.

NemiM [27]3 years ago

7 0

1. Define Newtons second law of motion (this will help put things into perspective)

2.Get the mass of the object (in this case 75 kg)

3.The net force acting on the object...find it (in this case 500 N)

4.Change the equation to F=ma (500=75a)

5.Divide both sides by 75 and that is the acceleration.

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A boy throws a stone straight upward with an initial

Dominik [7]

Answer:

14.8m

Explanation:

Given parameters:

Initial speed = 17m/s

Unknown:

Maximum height = ?

Solution:

At the maximum height, the final speed will be 0m/s;

We use of the kinematics equation to solve this problem.

V² = U² - 2gH

V is the final velocity

U is the initial velocity

g is the acceleration due to gravity

H is the height

0² = 17² - (2 x 9.8 x h )

0 = 289 - (9.6h)

-289 = -19.6h

h = 14.8m

5 0

2 years ago

A.complete the following sentences with your own result or condition

ludmilkaskok [199]

1. If I hadn't bought that car yesterday.

2. I would have never bought the tickets for today..

3. She wouldn't have gotten suspended.

4. We could've worked on our project.

5. but I was in the shower.

6 0

3 years ago

What is an instrument commonly used to measure wind speed?

xeze [42]

The correct answer is letter D. Anemometer. It is a device that is used to measure wind speed. It is a very common weather station instrument and is available to use and to make. Anemos, from the greek word that means wind.

5 0

2 years ago

Read 2 more answers

Assume that a pendulum used to drive a grandfather clock has a length L0=1.00m and a mass M at temperature T=20.00°C. It can be

Sedaia [141]

Answer:

The period will change a 0,036 % relative to its initial state

Explanation:

When the rod expands by heat its moment of inertia increases, but since there was no applied rotational force to the pendulum , the angular momentum remains constant. In other words:

ζ= Δ(Iω)/Δt, where ζ is the applied torque, I is moment of inertia, ω is angular velocity and t is time.

since there was no torque ( no rotational force applied)

ζ=0 → Δ(Iω)=0 → I₂ω₂ -I₁ω₁ = 0 → I₁ω₁ = I₂ω₂

thus

I₂/I₁ =ω₁/ω₂ , (2) represents final state and (1) initial state

we know also that ω=2π/T , where T is the period of the pendulum

I₂/I₁ =ω₁/ω₂ = (2π/T₁)/(2π/T₂)= T₂/T₁

Therefore to calculate the change in the period we have to calculate the moments of inertia. Looking at tables, can be found that the moment of inertia of a rod that rotates around an end is

I = 1/3 ML²

Therefore since the mass M is the same before and after the expansion

I₁ = 1/3 ML₁² , I₂ = 1/3 ML₂² → I₂/I₁ = (1/3 ML₂²)/(1/3 ML₁²)= L₂²/L₁²= (L₂/L₁)²

since

L₂= L₁ (1+αΔT) , L₂/L₁=1+αΔT , where ΔT is the change in temperature

now putting all together

T₂/T₁=I₂/I₁=(L₂/L₁)² = (1+αΔT) ²

finally

%change in period =(T₂-T₁)/T₁ = T₂/T₁ - 1 = (1+αΔT) ² -1

%change in period =(1+αΔT) ² -1 =[ 1+18×10⁻⁶ °C⁻¹ *10 °C]² -1 = 3,6 ×10⁻⁴ = 3,6 ×10⁻² % = 0,036 %

4 0

3 years ago

A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the

lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

$l^2 = x^2 + y^2-----(1)$