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2 years ago

What is the weight of a 15kg dog on earth?

1 answer:

3 0

In physics the standard unit of weight is Newton, and the standard unit of mass is the kilogram. On Earth, a 1 kg object weighs 9.8 N, so to find the weight of an object in N simply multiply the mass by 9.8 N. Or, to find the mass in kg, divide the weight by 9.8 N.

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What role did gravity play in the formation of the planets?

joja [24]

Your answer would be D.
If an object has mass, it has gravity, and the more mass it has, the stronger its gravity. During the formation of planets, essentially, various matter and elements pulled and fused together (because of the gravity), forming planetesimals.

3 0

3 years ago

Read 2 more answers

A car is pushed with a force of 450 N for 19.4 seconds. What impulse was applied to the car?

harina [27]

Answer:

impulse = 8820 kg· $\frac{m}{s}$ or 8820 N·s

Explanation:

Impulse J is equal to the average force $F_{av}$ multiplied by the elapsed time Δt or in equation form, J = $F_{av}$ Δt

As long as your force of 450 N is constant then that value is your average force $F_{av}$ and your elapsed time is 19.4 seconds.

Multiply these values.

You will get an impulse of 8820 kg· $\frac{m}{s}$ or 8820 N·s.

6 0

2 years ago

Fellow student of mathematical bent tells you that the wave function of a traveling wave on a thin rope is y(x,t)= 2.20 mm cos[(

bezimeni [28]

Answer

given,

y(x,t)= 2.20 mm cos[( 7.02 rad/m )x+( 743 rad/s )t]

length of the rope = 1.33 m

mass of the rope = 3.31 g

comparing the given equation from the general wave equation

y(x,t)= A cos[k x+ω t]

A is amplitude

now on comparing

a) Amplitude = 2.20 mm

b) frequency =

$f = \dfrac{\omega}{2\pi}$

$f = \dfrac{743}{2\pi}$

f = 118.25 Hz

c) wavelength

$k= \dfrac{2\pi}{\lambda}$

$\lambda= \dfrac{2\pi}{k}$

$\lambda= \dfrac{2\pi}{7.02}$

$\lambda= 0.895\ m$

d) speed

$v = \dfrac{\omega}{k}$

$v = \dfrac{743}{7.02}$

v = 105.84 m/s

e) direction of the motion will be in negative x-direction

f) tension

$T = \dfrac{v^2\ m}{L}$

$T = \dfrac{(105.84)^2\times 3.31 \times 10^{-3}}{1.33}$

T = 27.87 N

g) Power transmitted by the wave

$P = \dfrac{1}{2}m\ v \omega^2\ A^2$

$P = \dfrac{1}{2}\times 0.00331\times 105.84\times 743^2\ 0.0022^2$

P = 0.438 W

5 0

2 years ago

how does the electric force between two charged particles change if the distance between them is increased by a factor of 3? a.

Aleksandr [31]

how does the electric force between two charged particles change if the distance between them is increased by a factor of 3?

a. it is reduced by a factor of 3

4 0

2 years ago

Say you want to make a sling by swinging a mass M of 2.3 kg in a horizontal circle of radius 0.034 m, using a string of length 0

ycow [4]

Answer:

T = 764.41 N

Explanation:

In this case the tension of the string is determined by the centripetal force. The formula to calculate the centripetal force is given by:

$F_c=m\frac{v^2}{r}$ (1)

m: mass object = 2.3 kg

r: radius of the circular orbit = 0.034 m

v: tangential speed of the object

However, it is necessary to calculate the velocity v first. To find v you use the formula for the kinetic energy:

$K=\frac{1}{2}mv^2$

You have the value of the kinetic energy (13.0 J), then, you replace the values of K and m, and solve for v^2:

$v^2=\frac{2K}{m}=\frac{2(13.0J)}{2.3kg}=11.3\frac{m^2}{s^2}$

you replace this value of v in the equation (1). Also, you replace the values of r and m:

$F_c=(2.3kg)(\frac{11.3m^2/s^2}{0.034})=764.41N$

hence, the tension in the string must be T = Fc = 764.41 N

5 0

2 years ago