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2 years ago

WORTH 50 POINTSSSS!!!!!!!! don't lie either if you do I will report your answer and get my points back idc !!!!!!

Physics

1 answer:

BARSIC [14]2 years ago

3 0

Answer:

Explanation:

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Two forces are applied on a body. One produces a force of 480-N directly forward while the other gives a 513-N force at 32.4-deg

n200080 [17]

Answer:

F = (913.14 , 274.87 )

|F| = 953.61 direction 16.71°

Explanation:

To calculate the resultant force you take into account both x and y component of the implied forces:

$\Sigma F_x=480N+513Ncos(32.4\°)=913.14N\\\\\Sigma F_y=513sin(32.4\°)=274.87N$

Thus, the net force over the body is:

$F=(913.14N)\hat{i}+(274.87N)\hat{j}$

Next, you calculate the magnitude of the force:

$F=\sqrt{(913.14N)+(274.87N)^2}=953.61N$

and the direction is:

$\theta=tan^{-1}(\frac{274.14N}{913.14N})=16.71\°$

7 0

3 years ago

Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum

cestrela7 [59]

Answer:

$T_{2}=278.80 K$

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

$(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$ .

Now, let's use the ideal gas equation to the initial and the final state:

$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$

Let's recall that the term nR is a constant. That is why we can match these equations.

We can find a relation between the volumes of the initial and the final state.

$\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}$

Combining this equation with the first equation we have:

$(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$

$(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}$

Now, we just need to solve this equation for T₂.

$T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}$

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

$p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40$

Finally, T2 will be:

$T_{2}=278.80 K$

6 0

3 years ago

What characterizes static stretching? A. having a partner hold limbs in a stretch position B. assuming and holding a stretch pos

balandron [24]

Answer:

It is B

Trust me it is B

7 0

2 years ago

Read 2 more answers

A person travels by car from one city to another with differem constan1 speeds between pairs of cities. She drives for 30.0 min

Montano1993 [528]

Answer:

a.52.9 km/h

b.90 km

Explanation:

We are given that

$v_1=89km/h$

$t_1=30min$

$v_2=100km/h$

$t_2=12min$

$v_3=40km/h$

$t_3=45 min$

Time spend on eating lunch and buying ga=15 min.

a.Total time=30+12+45+15=102 minute= $\frac{102}{60}=1.7 hour$

1 hour=60 minutes

Distance= $speed\times time$

$d_1=v_1\times t_1=80\times\frac{30}{60}=40km$

$d_2=100\times \frac{12}{60}=20 km$

$d_3=40\times \frac{45}{60}=30 km$

Total distance= $d_1+d_2+d_3=40+20+30=90km$

Average speed= $\frac{total\;speed}{total\;time}$

Using the formula

Average speed= $\frac{90}{1.7}=52.9Km/h$

b.Total distance between the initial and final city lies along the route=90 km

4 0

3 years ago

Certain insects can achieve seemingly impossible accelerations while jumping. the click beetle accelerates at an astonishing 400

hichkok12 [17]

(a) The launching velocity of the beetle is 6.4 m/s

(b) The time taken to achieve the speed for launch is 1.63 ms

(a) The beetle starts from rest and accelerates with an upward acceleration of 400 g and reaches its launching speed in a distance 0.53 cm. Here g is the acceleration due to gravity.

Use the equation of motion,

$v^2=u^2+2as$

Here, the initial velocity of the beetle is u, its final velocity is v, the acceleration of the beetle is a, and the beetle accelerates over a distance s.

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 0.52×10⁻²m for s.

$v^2=u^2+2as\\ = (0 m/s)^2+2 (400)(9.8 m/s^2)(0.52*10^-^2 m)\\ =40.768 (m/s)^2\\ v=6.385 m/s$

The launching speed of the beetle is 6.4 m/s.

(b) To determine the time t taken by the beetle for launching itself upwards is determined by using the equation of motion,

$v=u+at$

Substitute 0 m/s for u, 400 g for a, 9.8 m/s² for g and 6.385 m/s for v.

$v=u+at\\ 6.385 m/s = (0 m/s) +400(9.8 m/s^2)t\\ t = \frac{6.385 m/s}{3920 m/s^2} = 1.63*10^-^3s=1.63 ms$

The time taken by the beetle to launch itself upwards is 1.62 ms.

(c) After the beetle launches itself upwards, it is acted upon by the earth's gravitational force, which pulls it downwards towards the earth with an acceleration equal to the acceleration due to gravity g. Its velocity reduces and when it reaches the maximum height in its path upwards, its final velocity becomes equal to zero.

Use the equation of motion,

$v^2=u^2+2as$

Substitute 6.385 m/s for u, -9.8 m/s² for g and 0 m/s for v.

$v^2=u^2+2as\\ (0m/s)^2=(6.385 m/s)^2+2(-9.8m/s^2)s\\ s=\frac{(6.385 m/s)^2}{2(9.8m/s^2)} =2.08 m$

The beetle can jump to a height of 2.1 m

7 0

3 years ago