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3 years ago

5

How do i get a negative cube when i calculate density

2 answers:

fgiga [73]3 years ago

8 0

374848282!,!( rnanxjcifjejzxj

NNADVOKAT [17]3 years ago

8 0

Answer:

density should always be positive, you must have missed a sign somewhere

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in a cricket match there are 5000 spectators counted 10 by 10 the number of significant figure in the measurement will be

BabaBlast [244]

Answer:

$\boxed{3 \ Significant \ figures}$

Explanation:

Total spectators = 5000

Counted by the groups of ten, So at last the result will be:

=> 5000/10 = 500

Significant figures in 500 are 3

8 0

3 years ago

Pls help im begging you

Lostsunrise [7]

Answer:

I think it's TRUE because forces change an object's motion but dont quote me on it ok? Cause I'm not a 100 percent sure

7 0

3 years ago

A grocery cart of mass 16 kg is being pushed at a constant speed up a 12-degree ramp by a force of [{MathJax fullWidth='false' f

lawyer [7]

Answer:

If the cart is being pushed at a constant speed, then the acceleration in the direction of motion is zero. Hence, the force in the direction of the motion is zero, according to Newton's Second Law.

$\Sigma F_x = ma_x$

For simplicity, I will denote the direction along the inclined ramp as x-direction.

In the question the value of the force is not clearly given, so I will denote it as $F_P$

$\Sigma F_{net_x} = ma_x\\\Sigma F = F_{P}\cos(29^\circ) - mg\sin(12^\circ) = ma_x = 0\\F_{P}\cos(29^\circ) = mgsin(12^\circ)\\F_{P}\times 0.8746 = 16\times 9.8\times 0.2079\\F_{P} = 37.2740$

Here the angle between the applied force and the x-direction is 12° + 17° = 29°

The x-component of the weight of the cart is equal to sine component of the weight.

Since the cart is rolling on tires the kinetic friction does no work.

Work done by the applied force:

$W_{F_P} = F_P_x \cos(29^\circ)\times 7.5 = 244.5 ~J$

Work done by the weight of the cart:

$W_{mg} = -mg\sin(12^\circ)\times 7.5 = -16\times 9.8 \times 0.2079 \times 7.5 = - 244.5~J$

Since the x-component of the weight is in the -x-direction, its work is negative.

Conveniently, the total work done on the particle is zero, since its velocity is constant.

5 0

3 years ago

What is the diameter of a 12lb shot if the specific gravity is of the shot iron in the shot is 6.8, the density of fresh water 6

marysya [2.9K]

Answer:

The diameter is 0.378 ft.

Explanation:

Given that,

Mass of shot = 12 lb

Density of fresh water = 62.4 lb/ft

Specific gravity = 6.8

We need to calculate the volume of shot

$V = \dfrac{4}{3}\pi r^3\ ft^3$

The density of shot is

Using formula of density

$\rho = \dfrac{m}{V}$

Put the value into the formula

$\rho =\dfrac{12}{ \dfrac{4}{3}\pi r^3}$

We need to calculate the radius

Using formula of specific gravity

$specific\ gravity =\dfrac{density\ of\ shot}{dnsity\ of\ water}$

Put the value into the formula

$6.8=\dfrac{\dfrac{12}{\dfrac{4}{3}\pi r^3}}{62.4}$

$r^3=\dfrac{12}{\dfrac{4}{3}\pi\times6.8\times62.4}$

$r^3=0.0067514$

$r =(0.0067514)^{\frac{1}{3}}$

$r=0.1890\ ft$

The diameter will be

$d = 2\times r$

$d =2\times0.1890$

$d =0.378\ ft$

Hence, The diameter is 0.378 ft.

7 0

3 years ago

Find the magnitude of the torque that acts on the molecule when it is immersed in a uniform electric field of 6.19×105 N/C with

Ivan

Answer:

$\tau=5.81\times 10^5p\ N-m$

Explanation:

We have,

Electric field, $E=6.19\times 10^5\ N/C$

The electric dipole vector at an angle of 69.9 degrees from the direction of the field.

The torque acting on a molecule is given by :

$\tau=p\times E\\\\\tau=pE\sin\theta$

p is electric dipole moment

$\tau=p\times 6.19\times 10^{5}\times \sin (69.9)\\\\\tau=5.81\times 10^5p\ N-m$

So, the magnitude of the torque acting on the molecule is $5.81\times 10^5p\ N-m$ .

7 0

4 years ago