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3 years ago

Which of the following statements is "true" of imagined risks?

Physics

1 answer:

Leno4ka [110]3 years ago

4 0

I think its A- They're unaffected by positive self talk

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In September 2016 you purchased a 125 shares of Grainger stock for $224.84. It is now September 2019 and Grainger stock is curre

Temka [501]

Answer:

Gain in capital = $ 70.72

Explanation:

Given:

- The price of stocks when purchased P_o = $ 224.84

- The price of stocks when sold P_s = $ 295.56

Find:

what would be your capital gain (loss) on the sale, ignoring commissions

Solution:

- The capital gain or loss on the selling of stocks stems from the difference of buying and selling value of stocks. The original price of stock was P_o and the selling price would be P_s. The difference would be:

capital gain = P_s - P_o

capital gain = $295.56 - $224.84

capital gain = $ 70.72

- Hence, there would be a gain in capital if sold today for about $ 70.72.

6 0

3 years ago

When a car of mass 1167 kg accelerates from 10.0 m/s to some final speed, 4.00  10 5J of work are done. Find this final speed.

Akimi4 [234]

Mass=1167kg
Initial velocity=u=10m/s
Acceleration=a=4m/s^2
Work done=105J=W
Final velocity=v=?
Force=F
Distance=d

Apply Newton's second law

$\\ \tt\hookrightarrow F=ma$

$\\ \tt\hookrightarrow F=1167(4)=4668N$

Now

$\\ \tt\hookrightarrow W=Fd$

$\\ \tt\hookrightarrow d=\dfrac{W}{F}$

$\\ \tt\hookrightarrow d=\dfrac{105}{4668}$

$\\ \tt\hookrightarrow d=0.022m$

Now

d be s

According to third equation of kinematics

$\\ \tt\hookrightarrow v^2=u^2+2as$

$\\ \tt\hookrightarrow v^2=10^2+2(4)(0.022)$

$\\ \tt\hookrightarrow v^2=100+8(0.022)$

$\\ \tt\hookrightarrow v^2=100+0.176$

$\\ \tt\hookrightarrow v^2=100.176$

$\\ \tt\hookrightarrow v=10.001m/s$

8 0

2 years ago

Which of the following is evidence of Earth's age provided by fossils?

Paladinen [302]

I think the answer is Fossils formed gradually over time.

7 0

3 years ago

Read 2 more answers

What is the final temperature when a 3.0 kg gold bar at 99 0C is dropped into 0.22 kg of water at 25oC?

liq [111]

Answer:

46.9 C

Explanation:

The heat released by the gold bar is equal to the heat absorbed by the water:

$m_g C_g (T_g-T_f)=m_w C_w (T_f-T_w)$

where:

$m_g = 3.0 kg$ is the mass of the gold bar

$C_g=129 J/kg C$ is the specific heat of gold

$T_g=99 C$ is the initial temperature of the gold bar

$m_w = 0.22 kg$ is the mass of the water

$C_w=4186 J/kg C$ is the specific heat of water

$T_w=25 C$ is the initial temperature of the water

$T_f$ is the final temperature of both gold and water at equilibrium

We can re-arrange the formula and solve for T_f, so we find:

$m_g C_g T_g -m_g C_g T_f = m_w C_w T_f - m_w C_w T_w\\m_g C_g T_g +m_w C_w T_w= m_w C_w T_f +m_g C_g T_f \\T_f=\frac{m_g C_g T_g +m_w C_w T_w}{m_w C_w + m_g C_g}=\\=\frac{(3.0)(129)(99)+(0.22)(4186)(25)}{(0.22)(4186)+(3.0)(129)}=\frac{38313+23023}{921+387}=\frac{61336}{1308}=46.9 C$

5 0

3 years ago

You are working with a 3 T MRI scanner with maximum gradient strength of 40 mT/m in all 3 axes (x,y,z). Your acquisition uses ma

inn [45]

Answer:

Rotational frequency=ω/2 $\pi$

Explanation:

Hence rotational frequency will be = 13.55 kHz

7 0

2 years ago