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3 years ago

Lillle is running. She increases her initial speed of 30 km/h to 40 km/h so she

Physics

2 answers:

Alex777 [14]3 years ago

6 0

Answer

$200km {h}^{ - 2}$

Explanation

$Acceleration = \frac{final \: \: \: velocity - initial \: \: velocity }{time} \\ = \frac{(40 - 30)km {h}^{ - 1} }{0.05h} \\ = \frac{10}{0.05} \\ = 200km {h}^{ - 2}$

Hope this helps you.

Let me know if you have any other questions :-):-)

Kaylis [27]3 years ago

5 0

initial velocity=u=30km/h
Final velocity=v=40km/h
Time=t=0.05h

$\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf\longmapsto Acceleration=\dfrac{40-30}{0.05}$

$\\ \sf\longmapsto Acceleration=\dfrac{10}{0.05}$

$\\ \sf\longmapsto Acceleration=200km/h^2$

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If a car moves with a uniform speed of 2m/s in a circle of radius 0.4m, what is its angular speed?

REY [17]

Answer:

The car's angular speed is $\frac{rad}{s}$ .

Explanation:

Angular velocity is usually measured with $\frac{radians}{sec}$ , so I'm going to use that to answer your question.

The relationship between tangential velocity and angular velocity (ω) is given by:

$V = \omega R$

Using the values from the question, we get:

$2 \frac{m}{s} = \omega (0.4m)$

$\omega = 5 \frac{rad}{s}$

Therefore, the car's angular speed is $5 \frac{rad}{s}$ .

Hope this helped!

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3 years ago

Enter an expression, in terms of defined quantities and g, for the force that the scale under left pillar shows

ella [17]

The expression, in terms of defined quantities and g is therefore Fu =((mg/2) +2 mp) g

<h3>What is a Scale?</h3>

This can be defined as a balance or any of various other instruments or devices for weighing.

The expression in terms of defined quantities and g, for the force that the scale under left pillar shows that Fu =((mg/2) +2 mp) g .

Read more about Force here brainly.com/question/4515354

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1 year ago

A thin, metallic spherical shell of radius 0.347 m0.347 m has a total charge of 7.53×10−6 C7.53×10−6 C placed on it. A point cha

USPshnik [31]

Answer:

E = 12640.78 N/C

Explanation:

In order to calculate the electric field you can use the Gaussian theorem.

Thus, you have:

$\Phi_E=\frac{Q}{\epsilon_o}$

ФE: electric flux trough the Gaussian surface

Q: net charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85*10^-12 C^2/Nm^2

If you take the Gaussian surface as a spherical surface, with radius r, the electric field is parallel to the surface anywhere. Then, you have:

$\Phi_E=EA=E(4\pi r^2)=\frac{Q}{\epsilon_o}\\\\E=\frac{Q}{4\pi \epsilon_o r^2}$

r can be taken as the distance in which you want to calculate the electric field, that is, 0.795m

Next, you replace the values of the parameters in the last expression, by taking into account that the net charge inside the Gaussian surface is:

$Q=7.53*10^{-6}C+3.65*10^{-6}C=1.115*10^{-5}C$

Finally, you obtain for E:

$E=\frac{1.118*10^{-5}C}{4\pi (8.85*10^{-12C^2/Nm^2})(0.795m)^2}=12640.78\frac{N}{C}$

hence, the electric field at 0.795m from the center of the spherical shell is 12640.78 N/C

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3 years ago

Radio telescopes are said to "listen" to space because...

almond37 [142]

Answer:

2 they detect radio waves

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3 years ago

A 87.0 kg astronaut is working on the engines of a spaceship that is drifting through space with a constant velocity. The astron

Ket [755]

Answer:

259.62521 seconds

Explanation:

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Here, the linear momentum is conserved

$m_1v_1=m_2v_2\\\Rightarrow v_1=\frac{m_2v_2}{m_1}\\\Rightarrow v_1=\frac{0.57\times 22.4}{87}\\\Rightarrow v_1=0.14675\ m/s$

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$Time=\frac{38.1}{0.14675}=259.62521\ s$

The time taken to reach the ship is 259.62521 seconds

4 0

3 years ago