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2 years ago

The weight of a body is 147N. what is its mass?

Physics

1 answer:

barxatty [35]2 years ago

5 0

To find the mass, with only the weight:

⇒ must consider the relationship between the mass and weight

⇒ (in other words) we must find the equation that has both the

mass and weight

Based on our physics knowledge, we know:

$Weight=mass*gravitational_. acceleration$

Weight: 147N
Gravitational Acceleration: 9.8 m/s²

Now let's plug in the values, and solve:

$147N = mass*9.8_.m/s^2\\mass = 15_.kg$

Answer: 15 kg

Hope that helps!

*as a note, if you use the gravitational acceleration as 10m/s², then the answer would be 14.7 kg

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Boyles law squeezing a balloon is one way to burst it. Why? ...?

KiRa [710]

According to Boyle's Law, volume is inversely proportional to pressure. It means if the volume of a gas goes up the pressure goes down and if the volume of the gas goes up the pressure goes down. When the pressure of air inside the inflated balloon is more than the atmospheric pressure outside the balloon. And also when the density inside is greater than the density outside. The molecules inside the balloon move and bang around the inner walls which produces force, which provides the pressure of an enclosed air.

6 0

3 years ago

An electron is in motion at 4.0 × 106 m/s horizontally when it enters a region of space between two parallel plates, as shown, s

max2010maxim [7]

Answer:

xmax = 9.5cm

Explanation:

In this case, the trajectory described by the electron, when it enters in the region between the parallel plates, is a semi parabolic trajectory.

In order to find the horizontal distance traveled by the electron you first calculate the vertical acceleration of the electron.

You use the Newton second law and the electric force on the electron:

$F_e=qE=ma$ (1)

q: charge of the electron = 1.6*10^-19 C

m: mass of the electron = 9.1*10-31 kg

E: magnitude of the electric field = 4.0*10^2N/C

You solve the equation (1) for a:

$a=\frac{qE}{m}=\frac{(1.6*10^{-19}C)(4.0*10^2N/C)}{9.1*10^{-31}kg}=7.03*10^{13}\frac{m}{s^2}$

Next, you use the following formula for the maximum horizontal distance reached by an object, with semi parabolic motion at a height of d:

$x_{max}=v_o\sqrt{\frac{2d}{a}}$ (2)

Here, the height d is the distance between the plates d = 2.0cm = 0.02m

vo: initial velocity of the electron = 4.0*10^6m/s

You replace the values of the parameters in the equation (2):

$x_{max}=(4.0*10^6m/s)\sqrt{\frac{2(0.02m)}{7.03*10^{13}m/s^2}}\\\\x_{max}=0.095m=9.5cm$

The horizontal distance traveled by the electron is 9.5cm

4 0

3 years ago

What do these letters stand for P=mv

victus00 [196]

Answer:

The equation for momentum of a piece of matter.

In either case, the momentum would be less than a linebacker hitting you at full speed. The equation for momentum is written: p = mv where p stands for momentum. That is, mass times velocity equals momentum.

Explanation:

Hope This Helps

Have A Great Day

4 0

2 years ago

A tugboat tows a ship at a constant velocity. The tow harness consists of a single tow cable attached to the tugboat at point A

Y_Kistochka [10]

Answer:

The tensions in $T_{BC}$ is approximately 4,934.2 lb and the tension in $T_{BD}$ is approximately 6,035.7 lb

Explanation:

The given information are;

The angle formed by the two rope segments are;

The angle, Φ, formed by rope segment BC with the line AB extended to the center (midpoint) of the ship = 26.0°

The angle, θ, formed by rope segment BD with the line AB extended to the center (midpoint) of the ship = 21.0°

Therefore, we have;

The tension in rope segment BC = $T_{BC}$

The tension in rope segment BD = $T_{BD}$

The tension in rope segment AB = $T_{AB}$ = Pulling force of tugboat = 1200 lb

By resolution of forces acting along the line A_F gives;

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = $T_{AB}$ = 1200 lb

$T_{BC}$ × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb............(1)

Similarly, we have for equilibrium, the sum of the forces acting perpendicular to tow cable = 0, therefore, we have;

$T_{BC}$ × sin(26.0°) + $T_{BD}$ × sin(21.0°) = 0...........................(2)

Which gives;

$T_{BC}$ × sin(26.0°) = - $T_{BD}$ × sin(21.0°)

$T_{BC}$ = - $T_{BD}$ × sin(21.0°)/(sin(26.0°)) ≈ - $T_{BD}$ × 0.8175

Substituting the value of, $T_{BC}$ , in equation (1), gives;

- $T_{BD}$ × 0.8175 × cos(26.0°) + $T_{BD}$ × cos(21.0°) = 1200 lb

- $T_{BD}$ × 0.7348 + $T_{BD}$ ×0.9336 = 1200 lb

$T_{BD}$ ×0.1988 = 1200 lb

$T_{BD}$ ≈ 1200 lb/0.1988 = 6,035.6938 lb

$T_{BD}$ ≈ 6,035.6938 lb

$T_{BC}$ ≈ - $T_{BD}$ × 0.8175 = 6,035.6938 × 0.8175 = -4934.1733 lb

$T_{BC}$ ≈ -4934.1733 lb

From which we have;

The tensions in $T_{BC}$ ≈ -4934.2 lb and $T_{BD}$ ≈ 6,035.7 lb.

8 0

3 years ago

Which of the following are steps for balancing chemical equations?

AveGali [126]

Answer:

C. Recheck the numbers of each atom on each side of the equation

to make sure the sides are equal.

D. Choose coefficients that will balance the equation

Explanation:

In balancing of chemical equation, the number of atoms on both sides must be equal in adherence to the law of conservation of mass.

Using the method of inspection, the equation is first observed to know the relationship between the combining atoms and the resulting ones.

After observing the reaction, put a coefficient that will balance the equation. Then recheck the number of each atom on both side of the equation. One can repeat the process till the equation is balanced.

5 0

3 years ago