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dedylja [7]
2 years ago
12

How much force (in Newtons) is needed to accelerate a 160 kg object at a rate of 4 m/s2?

Physics
1 answer:
Klio2033 [76]2 years ago
3 0

Answer:

C. 640

Explanation:

The minimum force would occur in a frictionless environment and act parallel to any abutting surface.

F = ma = 160(4) = 640 N

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A diesel engine lifts the 225 kg hammer of a pile driver 20 m in 5 seconds. How much work is done on
ladessa [460]

Answer:

a. Workdone = 44100 Joules

b. Power = 8820 Watts.

Explanation:

Given the following data:

Mass = 225kg

Distance = 20m

Time = 5 seconds

To find the workdone;

Workdone = force * distance

But force = mg

We know that acceleration due to gravity is equal to 9.8m/s²

Force = 225*9.8 = 2205N

Substituting the values into the equation, we have;

Workdone = 2205 * 20

Workdone = 44100 Joules

b. To find the power;

Power = workdone/time

Power = 44100/5

Power = 8820 Watts.

5 0
3 years ago
Physical properties used to measure stars include
V125BC [204]
The answer is D. Mass, brightness and size are used to measure the physical properties of the stars. One is the brightness it is a combination of the intrinsic brightness of a star. Second is the  size is calculated using the Stephan's Law. And the third is the mass, astronomers use the Stellar mass to describe the mass of the stars. 
5 0
3 years ago
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The diagram below shows a section of a roller coaster track. Five points on the track are labeled. Between which
arlik [135]

Answer: u and v. Trust me I took the test and i was right

7 0
3 years ago
Find the density of a planet with a radius of 8000 m if the gravitational acceleration for the planet, gp, has the same magnitud
Naya [18.7K]

Answer:

Density = 3 x 10⁻⁵ kg/m³

Explanation:

First, we will find the volume of the planet:

V = \frac{4}{3}\pi r^3\ (radius\ of\ sphere)\\\\V =   \frac{4}{3}\pi (8000\ m)^3\\\\V = 2.14\ x\ 10^{12}\ m^3

Now, we will use the expression for gravitational force to find the mass of the planet:

g = \frac{Gm}{r^2}\\\\m = \frac{gr^2}{G}

where,

m = mass = ?

g = acceleration due to gravity = 6.67 x 10⁻¹¹ m/s²

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ Nm²/kg²

r = radius = 8000 m

Therefore,

m = \frac{(6.67\ x\ 10^{-11}\ m/s^2)(8000\ m)^2}{6.67\ x\ 10^{-11}\ Nm^/kg^2}\\\\m = 6.4\ x\ 10^7\ kg

Therefore, the density will be:

Density = \frac{m}{V} = \frac{6.4\ x\ 10^7\ kg}{2.14\ x\ 10^{12}\ m^3}

<u>Density = 3 x 10⁻⁵ kg/m³</u>

4 0
3 years ago
How much kinetic energy does a system containing 3 moles of an ideal gas at 300 K possess? What is the heat capacity at constant
ElenaW [278]

Answer:

1. K.E = 11.2239 kJ ≈ 11.224 kJ

2. C_{V} = 37.413 JK^{- 1}

3. Q = 10.7749 kJ

Solution:

Now, the kinetic energy of an ideal gas per mole is given by:

K.E = \frac{3}{2}mRT

where

m = no. of moles = 3

R = Rydberg's constant = 8.314 J/mol.K

Temperature, T = 300 K

Therefore,

K.E = \frac{3}{2}\times 3\times 8.314\times 300

K.E = 11223.9 J = 11.2239 kJ ≈ 11.224 kJ

Now,

The heat capacity at constant volume is:

C_{V} = \frac{3}{2}mR

C_{V} = \frac{3}{2}\times 3\times 8.314 = 37.413 JK^{- 1}

Now,

Required heat transfer to raise the temperature by 15^{\circ} is:

Q = C_{V}\Delta T

\Delta T = 15^{\circ} = 273 + 15 = 288 K

Q = 37.413\times 288 = 10774.9 J = 10.7749 kJ

8 0
3 years ago
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