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2 years ago

Kenna has a mass of 30 kilograms. What's her weight in newtons? Assume that acceleration due to gravity is 9.8 N/kg.

Physics

1 answer:

astra-53 [7]2 years ago

5 0

Answer:

294 N

Explanation:

The weight of an object is the gravitational force exerted by the Earth on the object.

The direction of this force is always towards the Earth's centre, while the magnitude is given by:

$W=mg$

where

W is the weight

m is the mass of the object

g is the acceleration due to gravity

Here we have:

m = 30 kg is the mass of the object

g = 9.8 N/kg is the acceleration due to gravity

So, the weight of the object is

$W=(30)(9.8)=294 N$

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A labor push a crate through 12m and 720 J of work .The magnitude of force was

Harrizon [31]

Answer:

Explanation:

The magnitude of the force can be found by using the formula

$f = \frac{w}{d} \\$

w is the workdone

d is the distance

From the question we have

$f = \frac{720}{12} = 60 \\$

We have the final answer as

Hope this helps you

6 0

2 years ago

An object of mass 700700 kg is released from rest 20002000 m above the ground and allowed to fall under the influence of gravity

qwelly [4]

Answer:

59.503987 seconds

Explanation:

b = Proportionality constant = 50 Ns/m

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of object = 700 kg

We have the equation of velocity

$v(t)=\dfrac{mg}{b}+\left(V_0-\dfrac{mg}{b}\right)e^{\dfrac{bt}{m}}$

The equation of motion

$x(t)=\dfrac{mg}{b}+\dfrac{m}{b}\left(V_0-\dfrac{mg}{b}\right)(1-e^{\dfrac{bt}{m}})$

$x(t)=\dfrac{700\times 9.81}{50}+\dfrac{9.81}{50}\left(0-\dfrac{700\times 9.81}{50}\right)(1-e^{\dfrac{50t}{700}})$

when x(t)=2000

$2000=\dfrac{700\times 9.81}{50}+\dfrac{9.81}{50}\left(0-\dfrac{700\times 9.81}{50}\right)(1-e^{\dfrac{50t}{700}})\\\Rightarrow 2000\times \:50=\frac{700\times \:9.81}{50}\times \:50+\frac{9.81}{50}\left(0-\frac{700\times \:9.81}{50}\right)\left(1-e^{\frac{50t}{700}}\right)\times \:50\\\Rightarrow 6867-\frac{67365.27}{50}\left(1-e^{\frac{50t}{700}}\right)=100000\\\Rightarrow 50\left(-\frac{67365.27}{50}\left(1-e^{\frac{50t}{700}}\right)\right)=93133\times \:50\\\Rightarrow \frac{-67365.27\left(1-e^{\frac{50t}{700}}\right)}{-67365.27}=\frac{4656650}{-67365.27}\\\Rightarrow 1-e^{\frac{50t}{700}}=-69.12538\dots\\\Rightarrow -e^{\frac{50t}{700}}=-70.12538\dots\\\Rightarrow t=14\ln \left(70.12538\dots \right)\\\Rightarrow t=59.50398\ s$

The time taken is 59.503987 seconds

8 0

2 years ago

The weight of a metal bracelet is measured to be 0.10400 N in air and 0.08400 N when immersed in water. Find its density.

Anna007 [38]

Answer:

The density of the metal is 5200 kg/m³.

Explanation:

Given that,

Weight in air= 0.10400 N

Weight in water = 0.08400 N

We need to calculate the density of metal

Let $\rho_{m}$ be the density of metal and $\rho_{w}$ be the density of water is 1000kg/m³.

V is volume of solid.

The weight of metal in air is

$W =0.10400\ N$

$mg=0.10400$

$\rho V g=0.10400$

$Vg=\dfrac{0.10400}{\rho_{m}}$ .....(I)

The weight of metal in water is

Using buoyancy force

$F_{b}=0.10400-0.08400$

$F_{b}=0.02\ N$

We know that,

$F_{b}=\rho_{w} V g$ ....(I)

Put the value of $F_{b}$ in equation (I)

$\rho_{w} Vg=0.02$

Put the value of Vg in equation (II)

$\rho_{w}\times\dfrac{0.10400}{\rho_{m}}=0.02$

$1000\times\dfrac{0.10400}{0.02}=\rho_{m}$

$\rho_{m}=5200\ kg/m^3$

Hence, The density of the metal is 5200 kg/m³.

6 0

3 years ago

Particles 1 and 2 of charge q1 = q2 = +3.20 × 10−19 C are on a y axis at distance d = 17.0 cm from the origin. Particle 3 of cha

mel-nik [20]

Answer:

(a) 0.17 m

(b) 5.003 m

(d) 7.37 × $10^{-29}$ N

Explanation:

(a) The minimum value of $x$ will occur when q3 = 0 m or at origin and q1, q2 are at 0.17 m so the distance between q3 and q1, q2 is 0.17 m, therefore the <em>minimum value of x= 0.17 m</em>.

(b) The maximum value of x will occur when q3 = 5 m because it is said in the question that 5 is the maximum distance travelled by q3. To find the hypotenuse i.e. the distance between q3 and q1,q2, we use Pythagoras theorem.

$h^{2} = b^{2} + p^{2}$

$h^{2} = 5^{2} + 0.17^{2} \\h = \sqrt{} 25.03\\h= 5.002 m$

<em>Hence, the maximum distance is 5.002 m</em>

Minimum Distance = 0.17 m

For electrostatic force= $F=\frac{kq1q2}{x^{2} }$

$F=\frac{9 x 10^{9} x3.2x10^{-19}x 6.4x10^{-19} }{0.17^{2} }$

$F= 6.38$ × $10^{-26} N$

(d) For maximum magnitude, we use the maximum distance calculated in (b)

Maximum Distance = 5.002 m

Using the formula for electrostatic force again:

F = $\frac{9x10^{9}x3.2x10^{-19}x6.4x10^{-19} }{5.002^{2} } }$

F= 7.37× $10^{-29$ N

4 0

2 years ago

Arigid body must rotate about an axis in order for it to have angular momentum about that axis. True False

kompoz [17]

Answer:

False

Explanation:

Let's consider the definition of the angular momentum,

$\vec{L} = I \vec{\omega}$

where $I = \int\limits_m r^2 dm = \lim_{n \to \infty} \sum\limits_{i=1}^n m_i r_i^2$ is the moment of inertia for a rigid body. Now, this moment of inertia could change if we change the axis of rotation, because "r" is defined as the distance between the puntual mass and the nearest point on the axis of rotation, but still it's going to have some value. On the other hand,

$\vec{\omega} = \frac{\vec{r} \times \vec{v}}{r^2}$ so $\vec{\omega} \neq 0$ unless $\vec{r}$ ║ $\vec{v}$ .

In conclusion, a rigid body could rotate about certain axis, generating an angular momentum, but if you choose another axis, there could be some parts of the rigid body rotating around the new axis, especially if there is a projection of the old axis in the new one.

7 0

2 years ago