If the distance between two objects is tripled, what happens

What is the force felt by the 64-kg occupant of the car? Express your answer to two significant figures and include the appropri

Yakvenalex [24]

Answer:

The force is $-1.67\times10^{5}\ N$

Explanation:

Given that,

Mass of car = 64 kg

Suppose, a 1400-kg car that stops from 34 km/h on a distance of 1.7 cm.

We need to calculate the acceleration

Using formula of acceleration

$v^2-u^2=2as$

Where, v = final velocity

u = initial velocity

a = acceleration

s = distance

Put the value into the formula

$0^2-(34\times\dfrac{5}{18})^2=2\times a\times 1.7\times10^{-2}$

$a=\dfrac{(34\times\dfrac{5}{18})^2}{2\times1.7\times10^{-2}}$

$a=-2623.45\ m/s²$

We need to calculate the force

Using formula of force

$F=ma$

$F=64\times(-2623.45)$

$F=-1.67\times10^{5}\ N$

Negative sign shows the direction of the force is in the direction opposite to the initial velocity.

Hence, The force is $-1.67\times10^{5}\ N$

5 0

3 years ago

A 2 kilogram bowling ball is 2.5 meters off the ground on a post when it falls just before it reaches the ground it is traveling

s2008m [1.1K]

I need the answer too

7 0

3 years ago

Answers are ONLY allowed to be ionic, polar covalent, and nonpolar covalent.

Luba_88 [7]

A) It will be 2 covalent bonds
B) covalent bonds occur when there’s 2 atoms that share electrons. In this case by sharing the 2 pairs of valence electrons each atom has a total of 8 valence electrons

8 0

2 years ago

what type of energy appears when a gymnast jumps on to a spring board? Question is on energy stores!Thanks

Brrunno [24]

Hello There!

It is Spring potential energy. Also called Elastic potential energy.

Hope This Helps You!
Good Luck :)

- Hannah ❤

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;

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;

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

The change in heat energy for aluminum metal ;

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

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

3 years ago