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2 years ago

What is the speed of a car that travels 150 km in 3.00 hrs?

Physics

2 answers:

zhuklara [117]2 years ago

4 0

You would do distance divided by speed. So 150÷3, which would equal 5km per hour.

nalin [4]2 years ago

3 0

5km per hour. Hope this helps!

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How do you think that changing the mass of the pendulum bob will affect the period of the pendulum swing?

myrzilka [38]

(Mass does not affect the pendulum's swing. The longer the length of string, the farther the pendulum falls; and therefore, the longer the period, or back and forth swing of the pendulum. The greater the amplitude, or angle, the farther the pendulum falls; and therefore, the longer the period.)

5 0

2 years ago

If yall friend me ill give lots of points like this in the future

Ne4ueva [31]

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2 years ago

where σ(t) and σ(0) represents the time-dependent and initial (i.e., time =0) stresses, respectively, and t and τ denote elapsed

lesya [120]

Answer:

$E_r(6)=4.35614\ MPa$

Explanation:

$\epsilon$ = Strain = 0.49

$\sigma _0$ = 3.1 MPa

At t = Time = 32 s $\sigma$ = 0.41 MPa

$\tau$ = Time-independent constant

Stress relation with time

$\sigma=\sigma _0exp\left(-\frac{t}{\tau}\right)$

at t = 32 s

$0.41=3.1exp\left(-\frac{32}{\tau}\right)\\\Rightarrow exp\left(-\frac{32}{\tau}\right)=\frac{0.41}{3}\\\Rightarrow -\frac{32}{\tau}=ln\frac{0.41}{3}\\\Rightarrow \tau=-\frac{32}{ln\frac{0.41}{3}}\\\Rightarrow \tau=16.0787\ s$

The time independent constant is 16.0787 s

$E_{r}(t)=\frac{\sigma(t)}{\epsilon_0}$

At t = 6

$\\\Rightarrow E_{r}(6)=\frac{\sigma(6)}{\epsilon_0}$

From the first equation

$\sigma(t)=\sigma _0exp\left(-\frac{t}{\tau}\right)\\\Rightarrow \sigma(6)=3.1exp\left(-\frac{6}{16.0787}\right)\\\Rightarrow \sigma(6)=2.13451$

$E_r(6)=\frac{2.13451}{0.49}\\\Rightarrow E_r(6)=4.35614\ MPa$

$E_r(6)=4.35614\ MPa$

6 0

2 years ago

If 2.0×10^−4 C of charge passes a point in 2.0×10^−6 s , what is the rate of current flow?1.0×10^−10 A1.0×10^2 A4.0×10^−1 A4.0×1

Ipatiy [6.2K]

This problem is about the rate of the current. It's important to know that refers to the quotient between the electric charge and the time, that's the current rate.

$I=\frac{Q}{t}$

Where Q = 2.0×10^−4 C and t = 2.0×10^−6 s. Let's use these values to find I.

$\begin{gathered} I=\frac{2.0\times10^{-4}C}{2.0\times10^{-6}\sec } \\ I=1.0\times10^{-4-(-6)}A \\ I=1.0\times10^{-4+6}A \\ I=1.0\times10^2A \end{gathered}$

<em>As you can observe above, the division of the powers was solved by just subtracting their exponents.</em>

<em />

<h2>Therefore, the rate of the current flow is 1.0×10^2 A.</h2>

3 0

8 months ago

4. A 1,000 kg truck moving at 10 m/s runs into a concrete wall. It takes 0.5 seconds for the truck to conipietery

Maru [420]

Answer:

$\large \boxed{\text{h. 20 000 N}}$

Explanation:

Force is the change in momentum over time

F = Δp/Δt

1. Calculate the change in momentum

p₁ = mv₁ = 1000 kg × 10 m/s = 10 000 kg·m·s⁻¹

p₂ = 0

Δp = p₂ - p₁= (0 - 10 000) kg·m·s⁻¹ = -10 000 kg·m·s⁻¹

2. Calculate the force

$\begin{array}{rcl}F & = & \dfrac{\Delta p}{\Delta t}\\\\& = & \dfrac{-10 000 \text{ kg$\cdot$m$\cdot$ s}^{-1}}{\text{ 0.5 s}}\\\\& = & \textbf{-20 000 N}\\\end{array}\\\text{The negative sign shows that the force is exerted opposite to the direction of motion.}\\\text{The magnitude of the force is $\large \boxed{\textbf{20 000 N}}$}$

3 0

2 years ago

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