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nadezda [96]
3 years ago
11

What's the weight of a 30x30x50 cm body with the density of 1.8/cm cube?

Physics
1 answer:
grin007 [14]3 years ago
3 0

Answer:

The weight of the body, W = 793.8 m/s²

Explanation:

Given,

The volume of the body, v = 45,000 cm³

The density of the body, ρ = 1.8 g/cm³

The mass of the body is given by the formula,

                                  m = ρ x v

                                      = 1.8 g/cm³ x 45,000 cm³

                                      = 81,000 g

Hence, the mass of the body is m = 81 kg

The weight of the body,

                                           W = m x g

                                                = 81 kg x 9.8 m/s²

                                                = 793.8 m/s²

Hence, the weight of the body, W = 793.8 m/s²

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The glass in a window is 35 inches wide and 20 inches tall, and standard atmospheric pressure is 14.7 pounds per square inch. Wh
iogann1982 [59]

Answer:103 pounds

Explanation:

Given

width of window b=35 in.

height of window h=20 in.

standard atmospheric pressure P_{outside}=14.7 psi

Also P_{inside}=1.01P_{outside}

Thus Net Force on the window will be the algebraic sum of Force due to outside and inside Pressure .

F_{net}=(P_{inside}-P_{outside})\cdot A

F_{net}=P_{outside}(1.01-1)\times 35\times 20

F_{net}=14.7\times 0.01\times 35\times 20

F_{net}=102.9\ pounds\approx 103\ pounds            

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3 years ago
Hii can someone please doo this! 50 pointss.
andrey2020 [161]

Answer:

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Suppose your car is on a 5% grade, meaning that for every 100 m you travel along the road you raise or lower only 5 m in elevati
kvv77 [185]

Answer:

734.215N

Explanation:

First we calculate the angle that corresponds to a 5% slope using the Tan-1 function

\beta = tan-1(5%)=2.86

then we use the component that corresponds to the direction parallel to the road, additionally we must multiply by the gravity value to find the weight(g=9.81m/s^2)

Wx=M*g*sen(2.86)=1500kg*9.81*sen(2.86)=734.215N

8 0
3 years ago
Read 2 more answers
the total positive charge is QQQ = 1.62×10−6 CC , what is the magnitude of the electric field caused by this charge at point P,
balu736 [363]

Answer:

6.1 × 10^9 Nm-1

Explanation:

The electric field is given by

E= Kq/d^2

Where;

K= Coulombs constant = 9.0 × 10^9 C

q = magnitude of charge = 1.62×10−6 C

d = distance of separation = 1.53 mm = 1.55 × 10^-3 m

E= 9.0 × 10^9 × 1.62×10−6/(1.55 × 10^-3 )^2

E= 14.58 × 10^3/2.4 × 10^-6

E= 6.1 × 10^9 Nm-1

8 0
3 years ago
If a spring has a spring constant of 1.00 × 10^3 N/m, what is the restoring force when the mass has been displaced 20.0 cm?
never [62]

The spring has a spring constant of 1.00 * 10^3 N/m and the mass has been displaced 20.0 cm then the restoring force is 20000 N/m.

Explanation:

When a spring is stretched or compressed its length changes by an amount x from its equilibrium length then the restoring force is exerted.

spring constant is k = 1.00 * 10^3 N/m

mass is x = 20.0 cm

According to Hooke's law, To find restoring force,

    F = - kx

        = - 1.00 *10 ^3 * 20.0

     F = 20000 N/m

Thus, the spring has a spring constant of 1.00 * 10^3 N/m and the mass has been displaced 20.0 cm then the restoring force is 20000 N/m.

3 0
3 years ago
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