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3 years ago

Please help me this is timed . Find x if a = 3.0 m/s^2

Physics

1 answer:

storchak [24]3 years ago

5 0

Answer:

x = 50 N

Explanation:

Given that we have a net force, a mass, and acceleration, we can use the fundamental formula for force found in newton's second law which is F = m × a.

Given a mass of 150 kg, and an acceleration 3.0m/s². We can substitute these two values in our formula to calculate the magnitude of these forces or it's net force to identify the unknown force acting on our known force for this situation to work.

_______

F (Net force) = F2 (Second force which we are given) - F1 (First force) = m × a

m (mass which we are given) = 150 kg

a (acceleration which we are given) = 3.0m/s

________

So F = m × a → F2 - F1 = m × a →

500 - F1 = 150 × 3.0 → 500 - F1 = 450 →

-F1 = -50 → F1 = 50

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What is the atmospheric pressure and temperature at sea level in a standard<br> atmosphere?

Lady bird [3.3K]

Answer:

The tropospheric tabulation continues to 11,000 meters (36,089 ft), where the temperature has fallen to −56.5 °C (−69.7 °F), the pressure to 22,632 pascals (3.2825 psi), and the density to 0.3639 kilograms per cubic meter (0.02272 lb/cu ft). Between 11 km and 20 km, the temperature remains constant

Explanation:

Hope this helped, Have a wonderful day!!

4 0

3 years ago

a runner starts from rest and has an acceleration of 3 m/s^2. How fast is she running after 1.1 seconds

grigory [225]

Acceleration x time = velocity

Since you're given acceleration and time, just plug the values into the equation.

3 $\frac{m}{s^{2}}$ x 1.1 s = ?

Solve that equation, and remember your velocity should be in m/s.

8 0

3 years ago

50 grams of ice cubes at -15°C are used to chill a water at 30°C with mass mH20 = 200 g. Assume that the water is kept in a foam

Arada [10]

Answer : The final temperature is, $25.0^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ice = $2.09J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ice = 50 g

$m_2$ = mass of water = 200 g

$T_f$ = final temperature = ?

$T_1$ = initial temperature of ice = $-15^oC$

$T_2$ = initial temperature of water = $30^oC$

Now put all the given values in the above formula, we get:

$50g\times 2.09J/g^oC\times (T_f-(-15))^oC=-200g\times 4.184J/g^oC\times (T_f-30)^oC$

$T_f=25.0^oC$

Therefore, the final temperature is, $25.0^oC$

5 0

3 years ago

You are standing 2.5m directly in front of one of the two loudspeakers. They are 3.0m apart and both are playing a 686Hz tone in

ahrayia [7]

Answer:

distance from speaker is 17.87 m

Explanation:

given data

distance from loudspeaker = 2.5 m

distance between loudspeaker = 3.0 m

room temperature = 20c

wavelength f = 686Hz

to find out

what distances from the speaker

solution

we know sound velocity c = 331.5 + 0.6 × 20c = 343.5

so wavelength of sound λ = c / f

wavelength = 343.5 / 686 = 0.5 m

when the difference in distance of speaker destructive interference will be

d = λ/2 × (2n-1)

for n = 1, 2 3 4 ..

d = 0.5/2 × (2n-1)

d = 0.250 , 0.75 , 1.25 , 1.750............ for n = 1, 2 3 .............

for d = 0.250

side of triangle by hypotenuse of triangle are

$\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x1)$ = 0.250

0.5 x1 = 7.6875

x1 = 15.375 m

for d = 0.75

side of triangle by hypotenuse of triangle are

$\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x2)$ = 0.75

1.5 x2 = 4.6875

x2 = 3.125 m

for d = 1.250

side of triangle by hypotenuse of triangle are

$\sqrt{3^{2}+(2..5+x)^{2} } - (2.5 + x3)$ = 1.250

2.5 x2 = 1.1875

x3 = 0.475 m

for d = 1.750

x4 will be negative so we stop here

so the distance from speaker here is given below

distance = 2.5 + x

here x = 0.475 , 3.125 and 15.375 so

distance 1 = 2.5 + 0.475 = 2.975 m

distance 2 = 2.5 + 3.125 = 5.625 m

distance 3 = 2.5 + 15.375 = 17.875 m

final distance from speaker is 17.87 m

8 0

3 years ago

A point charge with charge q1 = 3.40 μC is held stationary at the origin. A second point charge with charge q2 = -4.90 μC moves

expeople1 [14]

Answer:

-0.79 J

Explanation:

We are given that

$q_1=3.4\mu C=3.4\times 10^{-6} C$

$1\mu C=10^{-6} C$

$q_2=-4.9\mu C=-4.9\times 10^{-6} C$

$x_1=0.125,y_1=0$

$x_2=0.280,y_2=0.235$

We have to find the work done by the electric force on the moving point charge.

$r_1=\sqrt{x^2_1+y^2_1}=\sqrt{(0.125)^2+0}=0.125$

$r_2=\sqrt{(0.280)^2+(0.235)^2}=0.366$

Work done, $W=kq_1q_2(\frac{1}{r_1}-\frac{1}{r_2})$

Where $k=9\times 10^9$

Using the formula

$W=9\times 10^9\times 3.4\times 10^{-6}\times(-4.9\times 10^{-6})(\frac{1}{0.125}-\frac{1}{0.366})$

$W=-0.79 J$

5 0

4 years ago