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4 years ago

a bus accelerates at -0.5 m/s^2 for 12.5 seconds if the bus had an initial speed of 31 /s what is its new speed

Physics

2 answers:

Arlecino [84]4 years ago

8 0

Answer:

as the other person said, the answer is 24.75 m/s

Explanation:

aleksandr82 [10.1K]4 years ago

3 0

Vf=v0+at so 31m/s-.5m/s^2*12.5= 24.75m/s is the new speed. the negative sign is showing a deceleration.

Hope this helps! Please rate if you like my answer! Thank you!!

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A projectile is launched from the surface of a planet (mass = 15 x 1024 kg, radius = R = 9.6 x 106 m). What minimum launch speed

Serjik [45]

Answer:

The minimum launch speed is 13366.40 m/s.

Explanation:

Given that,

Mass of planet $M=15\times10^{24}\ kg$

Radius $R= 9.6\times10^{6}\ m$

Height = 6R

We need to calculate the speed

Using relation between gravitational force and total energy

$-\dfrac{GMm}{6R+R}=\dfrac{1}{2}mv^2-\dfrac{GMm}{R}$

$\dfrac{}{}mv^2=\dfrac{GMm}{7R}-\dfrac{GMm}{R}$

$\dfrac{1}{2}mv^2=\dfrac{6GMm}{7R}$

$v^2=\dfrac{2\times6GM}{7R}$

Put the value into the formula

$v=\sqrt{\dfrac{2\times6\times6.67\times10^{-11}\times15\times10^{24}}{7\times9.6\times10^{6}}}$

$v=13366.40\ m/s$

Hence, The minimum launch speed is 13366.40 m/s.

6 0

3 years ago

Read 2 more answers

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;

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

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

Q = mcΔt

For copper metal, the quantity of heat transferred is calculated as;

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

The change in heat energy for copper metal;

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

For aluminum metal, the quantity of heat transferred is calculated as;

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