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3 years ago

7

When Sasha starts to ride a bike, it is very difficult for her to get going. When it comes time to stop, It's also hard to slow

down. Why is this?
A. unbalanced forces
B. Inertia
C. balanced forces

1 answer:

madreJ [45]3 years ago

3 0

Answer:

A, unbalanced forces

hope I helped (;

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Seeing how strong our gravitational pull is here on Earth, would it be possible to kill someone if you drop a penny off the Empi

nadezda [96]

That's an urban legend that's been around for a long time.
It was tested on "Mythbusters". The conclusion was that
it would sting pretty good, and maybe raise a lump, but
it would never penetrate the skull, and it couldn't kill.

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3 years ago

An information signal consists of a 25 Hz and a 75 Hz sine waves summed together. It is sampled at a frequency of 500 Hz, the hi

denpristay [2]

Answer:

Explanation:

An information contains

25Hz and 75Hz sine wave

Sample frequency is 500Hz

The analogy signal are generally

y(t) = Asin(2πx/λ - wt), w=2πf

y1(t)=Asin(2πx/λ - wt)

y1(t)=Asin(2πx/λ - 2π•25t)

y1(t)=Asin(2πx/λ - 50πt)

Similarly

y2(t)=Asin(2πx/λ - 150πt)

Using Nyquist theorem

Nyquist Theorem states that in order to adequately reproduce a signal it should be periodically sampled at a rate that is 2 times the highest frequency you wish to record.

From sampling

f(nyquist)=f(sample)/2

f(nyquist)=500/2

f(nyquist)=250Hz

From signal

The highest frequency is 150Hz

F(nyquist) = 2×F(highest)

f(nyquist)= 2×150

f(nyquist)= 300Hz

Sample per frequency Ns is given as

Ns=F(sample)/F(highest signal)

Ns=500/150

Ns=3.33sample/period

This is above nyquist rate of 2sample/period

So signal below 300Hz reproduced without aliasing.

The highest resulting frequency is 300Hz

6 0

3 years ago

If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her

Marysya12 [62]

Answer:

μ = 0.336

Explanation:

We will work on this exercise with the expressions of transactional and rotational equilibrium.

Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation

fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ = 0

fr sin θ - cos θ (W / 2 + 0,3 W_painter) = 0

fr = cotan θ (W / 2 + 0,3 W_painter)

Now let's write the equilibrium translation equation

X axis

F1 - fr = 0

F1 = fr

the friction force has the expression

fr = μ N

Y Axis

N - W - W_painter = 0

N = W + W_painter

we substitute

fr = μ (W + W_painter)

we substitute in the endowment equilibrium equation

μ (W + W_painter) = cotan θ (W / 2 + 0,3 W_painter)

μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)

we substitute the values they give

μ = cotan θ (12/2 + 0.3 55) / (12 + 55)

μ = cotan θ (22.5 / 67)

μ = cotan tea (0.336)

To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45

cotan 45 = 1 / tan 45 = 1

the result is

μ = 0.336

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4 years ago

Can someone help me with one through seven I will mark you the brainly

ki77a [65]

Answer:

1. F = M x A

2. Force

3. 2nd Law: Force

4. a, b, c (in order)

5. 3rd Law: Action and Reaction

6. b, c, a (in order)

7. 1st Law: Inertia

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3 years ago

Read 2 more answers

Help wit these questions someone.

romanna [79]

In series circuit, Req = R₁ + R₂ + R₃ + ···

In parallel circuit, $\frac{1}{Req} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} +...$

<h3>Q7.</h3>

total resistance in the upper branch = R₂ + R₃ = R₂ + 2

$\frac{1}{Req} = \frac{1}{R2+R3} + \frac{1}{R1}$

$\frac{1}{4} = \frac{1}{R2+2} +\frac{1}{6}$

R₂ + 2 = 12

R₂ = 10Ω

<h3>Q8.</h3>

$\frac{1}{Req} = \frac{1}{R2+R3} + \frac{1}{R1}$

$\frac{1}{Req} = \frac{1}{2+1} + \frac{1}{4}$

Req = 1.7Ω

7 0

3 years ago