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sergey [27]
3 years ago
10

Assuming a constant force if the mass of an object increases the acceleration of the object will

Physics
1 answer:
oksano4ka [1.4K]3 years ago
6 0

Answer:

The accelaration will decrease because force is inversly proportional to mass.

Explanation:

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A 14.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 63.0°-angle with the horizontal.
crimeas [40]

Answer:

Explanation:

Given:

length of ladder r_L = 14m

weight of ladder F_L = 490N

position of firefighter r_F = 3.8m

weight of firefighter F_F = 820N

angle of ladder \alpha = 63

Unknown:

force of the wall on the ladder F_W

force of friction on base of ladder F_R

normal force on base of ladder F_N

From the free body diagram of the sketch you get 3 equations:

F_x = ma_x = F_W - F_R = 0\\ F_y = ma_y = F_N - F_F - F_L = 0\\ \tau _P = \overrightarrow{r} \times \overrightarrow{F} = r_FF_Fcos\alpha + \frac{1}{2}r_LF_Lcos\alpha - r_LF_Wsin\alpha = 0

Solving the equations gives:

F_W = F_R\\ F_N = F_F + F_L\\ F_W = \frac{r_FF_F + 0.5r_LF_L}{r_L tan\alpha}

a)

F_R = 238N\\ F_N = 1310N

b)

F_R = \mu F_N\\ \mu = \frac{F_R}{F_N} \\ \mu = 0.3

c) Using the result from b and solving for r_F

\\ \mu = 0.15\\ F_R = \mu F_N\\ r_F = 2.4m

4 0
3 years ago
Witch of the following is the correct definition of convention
iogann1982 [59]

Answer:

A is the answer .(A) is correct

3 0
2 years ago
A dockworker applies a constant horizontal force of 90.0 N to a block of ice on a smooth horizontal floor. The frictional force
Oliga [24]

Answer:

The mass of the ice block is equal to 70.15 kg

Explanation:

The data for this exercise are as follows:

F=90 N

insignificant friction force

x=13 m

t=4.5 s

m=?

applying the equation of rectilinear motion we have:

x = xo + vot + at^2/2

where xo = initial distance =0

vo=initial velocity = 0

a is the acceleration

therefore the equation is:

x = at^2/2

Clearing a:

a=2x/t^2=(2x13)/(4.5^2)=1.283 m/s^2

we use Newton's second law to calculate the mass of the ice block:

F=ma

m=F/a = 90/1.283=70.15 kg

6 0
3 years ago
Read 2 more answers
The MAC is 58 inches, The CG limits are from 26% to 43% MAC. If the CG is found to be
eimsori [14]

By working with percentages, we want to see how many inches is the center of gravity out of the limits. We will find that the CG is 1.45 inches out of limits.

<h3>What are the limits?</h3>

First, we need to find the limits.

We know that the MAC is 58 inches, and the limits are from 26% to 43% MAC.

So if 58 in is the 100%, the 26% and 43% of that are:

  • 26% → (26%/100%)*58in = 0.26*58 in = 15.08 in
  • 43% → (43%/100%)*58in = 0.43*58 in = 24.94 in.

But we know that the CG is found to be 45.5% MAC, then it measures:

(45.5%/100%)*58in = 0.455*58in = 26.39 in

We need to compare it with the largest limit, so we get:

26.39 in - 24.94 in = 1.45 in

This means that the CG is 1.45 inches out of limits.

If you want to learn more about percentages, you can read:

brainly.com/question/14345924

6 0
2 years ago
Green light (λ = 518 nm) strikes a single slit at normal incidence. What width slit will produce a central maximum that is 3.00
MariettaO [177]

Answer:

6.9066 × 10⁻⁵ m

Explanation:

For constructive interference, the expression is:

d\times sin\theta=m\times \lambda

Where, m = 1, 2, .....

d is the distance between the slits.

The formula can be written as:

sin\theta=\frac {\lambda}{d}\times m ....1

The location of the bright fringe is determined by :

y=L\times tan\theta

Where, L is the distance between the slit and the screen.

For small angle , sin\theta=tan\theta

So,  

Formula becomes:

y=L\times sin\theta

Using 1, we get:

y=L\times \frac {\lambda}{d}\times m

Thus, the distance between the central maximum is 3.00 cm

First bright fringe , m = 1 occur at 3.00 / 2 = 1.50 cm

Since,

1 cm = 0.01 m

y = 0.0150 m

Given L = 2.00 m

λ = 518 nm

Since, 1 nm = 10⁻⁹ m

So,

λ = 518 × 10⁻⁹ m

Applying the formula as:

0.0150\ m=2.00\ m\times \frac {518\times 10^{-9}\ m}{d}\times 1

<u>⇒ d, distance between the slits = 6.9066 × 10⁻⁵ m</u>

7 0
3 years ago
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