The ozone the unearths surface from the dangerous radiation. This is an interaction between

the velocity of a car traveling in the positive direction decreases from 32 m/s to 24 m/s in 4 seconds. what is the average acce

anastassius [24]

Answer:

$a=-2m/sec^2$

Negative sign shows that velocity of the car is decreases at a constant rate

Explanation:

We have given velocity of the car is decreases from 32 m /sec to 24 m/sec in 4 second

So initial velocity of the car u = 32 m /sec

And finally car reaches to a velocity of 24 m/sec

Time taken to change in velocity = 4 sec

So final velocity v = 24 m/sec

From first equation of motion v = u+at

So $24=32+a\times 4$

$a=-2m/sec^2$

Negative sign shows that velocity of the car is decreases at a constant rate

6 0

3 years ago

Calculate the amount of heat transferred when 710 grams of water warms from an initial temperature of 4.0 ºC to a final temperat

tigry1 [53]

Answer:

Q = 62383.44 Joules

Explanation:

Given that,

Mass of water, m = 710 gm

Initial temperature of water, $T_i=4^{\circ} C$

Final temperature of water, $T_f=25^{\circ} C$

The specific heat capacity of liquid water is, $c=4.184\ J/g\ ^oC$

Heat transferred is given by :

$Q=mc(T_f-T_i)$

$Q=710\times 4.184\times (25-4)$

Q = 62383.44 Joules

So, the amount of heat transferred is 62383.44 Joules. Hence, this is the required solution.

3 0

3 years ago

A boat takes off from a dock at 2.5 m/s and speeds up at 4.2 m/s squared for six seconds how far has the most traveled

GaryK [48]

The boat's position $x$ relative to its starting point $x_0=0$ is determined by

$x=x_0+v_0t+\dfrac12at^2$

where $v_0$ is its initial velocity, $a$ is its acceleration, and $t$ is time. After $t=6\,\mathrm s$ , the boat has traveled

$x=\left(2.5\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(4.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2$

$\implies x=91\,\mathrm m$

3 0

3 years ago

Refrigerant-134a enters the expansion valve of a refrigeration system at 160 psia as a saturated liquid and leaves at 30 psia. D

KatRina [158]

Answer:

Temperature : 92.9 F

Internal Energy change: -2.53 Btu/lbm

Explanation:

As

mh1=mh2

h1=h2

In table A-11 through 13E

p2=120Psi, h1= 41.79 Btu/lbm,

u1=41.49

So T1=90.49 F

P2=20Psi

h2=h1= 41.79 Btu/lbm

T2= -2.43F

u2= 38.96 Btu/lbm

T2-T1 = 92.9 F

u2-u1 = -2.53 Btu/lbm

3 0

3 years ago

As shown in the diagram, two forces act on an object. The forces have magnitudes F1 = 5.7 N and F2 = 1.9 N. What third force wil

galina1969 [7]

Answer:

Second option 6.3 N at 162° counterclockwise from

F1->

Explanation:

Observe the attached image. We must calculate the sum of all the forces in the direction x and in the direction y and equal the sum of the forces to 0.

For the address x we have:

$-F_3sin(b) + F_1 = 0$

For the address and we have:

$-F_3cos(b) + F_2 = 0$

The forces $F_1$ and $F_2$ are known

$F_1 = 5.7\ N\\\\F_2 = 1.9\ N$

We have 2 unknowns ( $F_3$ and b) and we have 2 equations.

Now we clear $F_3$ from the second equation and introduce it into the first equation.

$F_3 = \frac{F_2}{cos (b)}$

Then

$-\frac{F_2}{cos (b)}sin(b)+F_1 = 0\\\\F_1 = \frac{F_2}{cos (b)}sin(b)\\\\F_1 = F_2tan(b)\\\\tan(b) = \frac{F_1}{F_2}\\\\tan(b) = \frac{5.7}{1.9}\\\\tan^{-1}(\frac{5.7}{1.9}) = b\\\\b= 72\°\\\\m = b +90\\\\\m= 162\°$

Then we find the value of $F_3$

$F_3 = \frac{F_1}{sin(b)}\\\\F_3 =\frac{5.7}{sin(72\°)}\\\\F_3 = 6.01 N$

Finally the answer is 6.3 N at 162° counterclockwise from

F1->

7 0

3 years ago