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blagie [28]
4 years ago
12

After driving a portion of the route, the taptap is fully loaded with a total of 25 people including the driver, with an average

mass of 62 kgkg per person. In addition, there are three 15-kgkg goats, five 3- kgkg chickens, and a total of 25 kgkg of bananas on their way to the market. Assume that the springs have somehow not yet compressed to their maximum amount. How much are the springs compressed
Physics
1 answer:
loris [4]4 years ago
3 0

Answer:

0.4455 m

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

Total mass is

m=62\times 25+15\times 3+5\times 3+25\\\Rightarrow m=1635\ kg

Here the spring constant is not given so let us assume it as k=36000\ N/m

Here, the forces are balanced

mg=kx\\\Rightarrow 1635\times 9.81=36000\times x\\\Rightarrow x=\dfrac{1635\times 9.81}{36000}\\\Rightarrow x=0.4455\ m

The springs are compressed by 0.4455 m

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Identical isolated conducting spheres 1 and 2 have equal charges and are separated by a distance that is large compared with the
scoundrel [369]

Answer:

F = 0.1575 N

Explanation:

When the third sphere touches the first sphere, the charge is distributed between both spheres, then now the first sphere has only half of his original charge.

In this moment then

Sphere one has a charge = Q/2

Sphere three has a charge = Q/2

Now when the third sphere touches the second sphere again the charge is distributed in a manner that both sphere has the same charge.

How the total charge is Q = Q/2 + Q = 3/2Q, when the spheres are separated each one has 3/4Q

Sphere two has a charge = 3/4Q

Sphere three has a charge = 3/4Q

The electrostatic force that acts on sphere 2 due to sphere 1 is:

F = \frac{kQ_{1}Q_{2} }{r^{2} }

F= \frac{K(Q/2)(3Q/4)}{r^{2} }

how \frac{KQ^{2} }{r^{2} } = 0.42

Then

F = \frac{0.42*3}{8}

F = 0.1575 N

3 0
3 years ago
Two wheels roll side-by-side without sliding, at the same speed. The radius of wheel 2 is one-half (1/2) the radius of wheel 1.
Lemur [1.5K]

Answer:

w'=(1/2)w

Explanation:

In order to calculate the angular velocity of the second wheel, you use the following formula:

\omega=\frac{v}{r}      (1)

v: speed of the wheel 1 = speed of the wheel 2

r: radius of the wheel 1

For the second wheel you have:

r'=2r

You replace this value of r' in the following equation:

\omega'=\frac{v}{r'}=\frac{v}{2r}=\frac{1}{2}\frac{v}{r}=\frac{1}{2}\omega\\\\\omega'=\frac{1}{2}\omega

The angular velocity of the second wheel is one half of the angular velocity of the first wheel

6 0
3 years ago
The precision of a balance is ±0.02 g. The accepted value for a measurement is 26.86 g. Which measurement is in the accepted ran
HACTEHA [7]

Answer:

Ben türküm sizi anlamıyom kb

hepiniz malsınız amk

7 0
3 years ago
Read 2 more answers
- A cannon of 2000 kg fires a shell of 10 kg at
fgiga [73]

1) -0.5 m/s

We can solve the first part of the problem by using the law of conservation of momentum. In fact, the total momentum of the cannon - shell system must be conserved.

Before the shot, both the cannon and the shell are at rest, so the total momentum is zero:

p=0

After the shot, the momentum is:

p=MV+mv

where

M = 2000 kg is the mass of the cannon

m = 10 kg is the mass of the shell

v = 100 m/s is the velocity of the shell (we take as positive the direction of motion of the shell)

V = ? is the velocity of the cannon

Since momentum is conserved, we can write

0=MV+mv

And solving for V, we find the velocity of the cannon:

V=-\frac{mv}{M}=-\frac{(10)(100)}{2000}=-0.5 m/s

where the negative sign indicates that the cannon moves in the direction opposite to the shell.

2) 0.5 m

The motion of the cannon is a uniformly accelerated motion, so we can solve this part by using suvat equation:

v^2-u^2=2as

where

v is the final velocity of the cannon

u = 0.5 m/s is the initial velocity of the cannon (now we take as positive the initial direction of motion of the cannon)

a=-0.25 m/s^2 is the deceleration of the cannon

s is the distance travelled by the cannon

The cannon will stop when v = 0; substituting and solving the equation for s, we find the minimum safe distance required to stop the cannon:

s=\frac{v^2-u^2}{2a}=\frac{0-0.5^2}{2(-0.25)}=0.5 m

7 0
4 years ago
They occupy the space surrounding the nucleus of the atom, they have a -1 charge
kipiarov [429]

Answer:

Electrons.

Explanation:

Look it up.

4 0
3 years ago
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