A man jogs at a speed of 1.6 m/s. His dog

Pesure is defined the force over a given area.

dezoksy [38]

Answer:yes absolutely

Explanation:

7 0

3 years ago

Read 2 more answers

a catcher "gives" with the ball when he catches a 0.196 kg baseball moving at 31 m/s. if he moves his glove a distance of 5.32 c

Aloiza [94]

Answer:

3540.5N

Explanation:

Step one:

given data

mass m= 0.196kg

speed v= 31m/s

distance r= 5.32cm = 0.0532m

Step two

The expression relating force, mass, velocity and distance is

F= mv^2/r

substitute we have

F=0.196*31^2/0.0532

F=0.196*961/0.0532

F=188.356/0.0532

F=3540.5N

6 0

3 years ago

A child in danger of drowning in a river is being carried downstream by a current that flows due south uniformly with a speed of

tia_tia [17]

Let the rescue boat starts at an angle theta with the North

now its velocity towards East is given as

$v_x = 24sin\theta$

$v_y = -24cos\theta + 3$

now in some time "t" it will catch the boy

so we will have

$t = \frac{0.5}{24sin\theta}$

also we have

$t = \frac{2}{-24cos\theta + 3}$

now we have

$\frac{2}{-24cos\theta + 3} = \frac{0.5}{24sin\theta}$

$4*24sin\theta = - 24cos\theta + 3$

$96 sin\theta + 24cos\theta = 3$

by solving above we got

$\theta = 164 degree$

3 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;

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