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3 years ago

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You drop a basketball (Mass = 0.5 kg) and a medicine ball (mass = 5 kg) from the window of the leaning tower of Pisa. Assuming t

here is very little air resistance , what will you observe ?

1 answer:

erastova [34]3 years ago

4 0

The medicine ball wall hit the ground faster, if that makes sense to you

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How many poles do you expect to see in your magnet? Look around your room. Make a list of 5 items that might be attracted to the

jeyben [28]

Answer:

1.1 Two poles: North and South Poles.

1.2 - Staple pin - Nail - Tip of my phone charger - Metal keys - Cloth Hanger

1.3 - Wooden bed cot - Plastic pen - Game pad - Wooden shelf - Paper - A T-shirt

1.4 Yes

1.5 No

5 0

3 years ago

For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par

Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

<em>initial temperature of metals, = </em> $T_m$ <em />
<em>initial temperature of water, = </em> $T_i$ <em> </em>
<em>specific heat capacity of copper, </em> $C_p$ <em> = 0.385 J/g.K</em>
<em>specific heat capacity of aluminum, </em> $C_A$ = 0.9 J/g.K
<em>both metals have equal mass = m</em>

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

<em>For</em><em> copper metal</em><em>, the quantity of heat transferred is calculated as</em>;

$Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w ) = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t$

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>copper metal</em>;

$\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t$

<em>For </em><em>aluminum metal</em><em>, the quantity of heat transferred is calculated as</em>;

$Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w) = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t$

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>aluminum metal </em><em>;</em>

$\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t$

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

Learn more here:brainly.com/question/15345295

6 0

2 years ago

If an IPS student weighs 115 lbs on Earth, what is their weight in Newtons?

Montano1993 [528]

Answer:

511.545 Newtons

Explanation:

So 1 pound=4.44822 Newton’s so 115 times 4.44822 is 511.5453, then round it to get 511.545 Newtons.

6 0

3 years ago

Read 2 more answers

A proton moving in the positive x direction with a speed of 9.9 105 m/s experiences zero magnetic force. When it moves in the po

Alex

Answer:

The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction

Explanation:

In order to calculate the magnitude and direction of the magnetic field, you take into account the following equation for the magnetic force on the proton:

$\vec{F_B}=q\vec{v}\ X\ \vec{B}$ (1)

v: speed of the proton = 9.9*10^5 m/s

q: charge of the proton = 1.6*10^-19C

B: magnetic field = ?

FB: magnetic force on the proton = 1.6*10^-13N

When the proton travels in the positive y direction (^j), you have that the proton experiences a force in the positive z direction (+^k). To obtain this direction of the magnetic force on the proton, it is necessary that the magnetic field points in the negative x direction, in fact, you have:

^j X (-^i) = -(-^k)=^k

To obtain the magnitude of the magnetic field you use:

$F_B=qvBsin90\°=qvB\\\\B=\frac{F_B}{qv}=\frac{1.6*10^{-13}N}{(1.6*10^{-19}C)(9.9*10^5m/s)}\\\\B=1.01T$

The magnitude of the magnetic field is 1.01T and its direction is in the negative x direction

8 0

3 years ago

When a user wants to participate in a PKI, what component does he or she need to obtain, and how does that happen?

Luba_88 [7]

Answer: certificate authority

Explanation: The user submits Identification data and certificate request to the registration authority (RA). The RA validates this information and sends the certificate request to the certificate Authority (CA)

5 0

3 years ago