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blagie [28]
3 years ago
7

A net force of magnitude 36 N gives an object of mass m1 an acceleration of 6.0 m/s^2. The same net force gives m1 and another o

bject of mass m2 fastened together an acceleration of 2.0 m/s^2. What acceleration will m2 experience if the same net force acts on it alone?
Physics
1 answer:
Jet001 [13]3 years ago
3 0
This is a beautiful problem to test whether a student actually understands
Newton's 2nd law of motion . . . Force = (mass) x (acceleration). 

That simple law is all you need to solve this problem, but you need to
use it a few times.

m₁ alone:
                     Force = (mass) x (acceleration)

                       36 N = ( m₁ ) x  (6 m/s²)

                         m₁ = (36 N) / (6 m/s²)   

                         m₁  =  6 kilograms .

m₁ and m₂ glued together:
                     Force = (mass) x (acceleration)

                       36 N = (6 kg + m₂) x (2 m/s²)

                     6 kg + m₂  =  (36 N) / (2 m/s²)  =  18 kilograms

                               m₂  =  12 kilograms .

m₂ alone:
                     Force = (mass) x (acceleration)

                       36 N = (12 kg) x (acceleration)

                     Acceleration = (36 N) / (12 kg)

                     Acceleration  =   3 m/s²


You might be interested in
3 a A motorcyclist starts from rest and reaches
andrew-mc [135]

The acceleration of the body is 2 m/s^2 while the deceleration is - 1.2 m/s^2.

<h3>What is the acceleration?</h3>

Let us recall that the acceleration is the change in the speed of a body with time. We have been told that the body accelerates for 3s and then decelerates to 2s. This implies that the total time that the object spent in motion is 5 s.

Thus;

v = u + at

v = final velocity

u = initial velocity

a = acceleration

t = time taken

v - u/t = a

a = 6 - 0/3

= 2 m/s^2

Again;

v - u/t = a

a = 0 - 6/5

a = - 1.2m/s^2

Learn more about acceleration:brainly.com/question/12550364

#SPJ1

<h3 />
7 0
1 year ago
Problem 3: Thermal expansionThe steel rod has the length 2 m and cross-section area 200 cm2at the room temperature 20◦C. Weapply
zalisa [80]

Answer:

a) 2.00024 m

b) 0.036%

c) 436.67°C

Explanation:

Given

Initial length = L₀ = 2 m

Initial cross sectional Area = A₀ = 200 cm² = 0.02 m²

We can obtain initial volume = V₀ = A₀L₀ = 0.02 × 2 = 0.04 m³

Initial Temperature = T₀ = 20°C

Coefficient of linear expansivity = α = (2 × 10⁻⁶) (°C)⁻¹

a) New length of the rod after heating to 80°C

Linear expansion is given as

ΔL = L₀ × α ×ΔT

ΔL = 2 × 2 × 10⁻⁶ × (80 - 20) = 0.00024 m = 0.24 mm

New length = old length + expansion = 2 + 0.00024 = 2.00024 m

b) The percentage of the volume change of the rod.

Volume expansion is given by

ΔV = V₀ × (3α) × ΔT

Volume expansivity ≈ 3 × (linear expansivity)

ΔV = 0.04 × (3×2×10⁻⁶) × (80 - 20) = 0.0000144 m³

Percentage change in volume = 100% × (ΔV/V₀) = 100% × (0.0000144/0.04) = 0.036%

c) The maximal temperature we can allow if the volume should not increase by more than half percent.

For a half percent increase in volume, the corresponding change in volume needs to be first calculated.

Percentage change in volume = 100% × (ΔV/V₀)

0.5 = 100% × (ΔV/0.04)

(ΔV/0.04) = 0.005

ΔV = 0.0002 m³

Then we now investigate the corresponding temperature that causes this.

ΔV = V₀ × (3α) × ΔT

0.0002 = 0.04 × (3×2×10⁻⁶) × ΔT

ΔT = (0.0002)/(0.04 × 3 × 2 × 10⁻⁶) = 416.67°C

Maximal temperature = T₀ + ΔT = 20 + 416.67 = 436.67°C

4 0
3 years ago
How will a current change if the resistance of a circuit remains constant while the voltage across the circuit decreases to half
beks73 [17]

Answer:

1. The current will drop to half of its original value.

Explanation:

The problem can be solved by using Ohm's law:

V=RI

where

V is the voltage across the circuit

R is the resistance of the circuit

I is the current

We can rewrite it as

I=\frac{V}{R}

In this problem, we have:

- the resistance of the circuit remains the same: R' = R

- the voltage is decreased to half of its original value: V'=\frac{V}{2}

So, the new current will be

I'=\frac{V'}{R'}=\frac{V/2}{R}=\frac{1}{2}\frac{V}{R}=\frac{I}{2}

so, the current will drop to half of its original value.

4 0
3 years ago
Based on the data given, in what direction will the car accelerate?
skad [1K]
Vertical forces:
There is a force of 579N acting upward, and a force of 579N
acting downward.
The vertical forces are balanced ... they add up to zero ...
so there's no vertical acceleration. 
Not up, not down.

Horizontal forces:
There is a force of 487N acting to the left, and a force of 632N
acting to the right.
The net horizontal force is

        (487-left + 632-right)  -  (632-right - 487-right) =  145N to the right.

The net force on the car is all to the right.
The car accelerates to the right.
7 0
3 years ago
If a fish is trying to capture an insect hovering above the surface of the water – how will it jump to catch it? Will it aim abo
Viktor [21]

Answer:

It will have aim at a point "below" the insect.

From the insect's point of view, the fish will appear to be shallower than it actually is because a ray of light from the insect to the fish will be bent "towards" the normal when the ray enters the water

5 0
2 years ago
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