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2 years ago

A force of 20N acts upon a body whose weight is 9.8 N. What is the mass of the body and how much is its acceleration?

Physics

2 answers:

zloy xaker [14]2 years ago

7 0

Answer:

Explanation:

The two triangles are similar what is the length of DE

Anton [14]2 years ago

5 0

Answer:

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____________ is an out made when a base runner, forced to run because another teammate must run to the base being occupied, cann

maxonik [38]

If I knew the answer I would help but I don’t know sorry

5 0

2 years ago

A child on a skateboard experiences a 75 N force with an acceleration of 1.5 m/s2.

cupoosta [38]

Answer:

The mass of the child + skateboard is 50 kg

Explanation:

In this problem, we can apply Newton's second law:

F = ma

where

F is the net force on a system

m is the mass of the system

a is the acceleration of the system

In this problem, our system is the child + the skateboard. The net force on them is

F = 75 N

and their acceleration is

$a=1.5 m/s^2$

So we can re-arrange the equation above to find their combined mass:

$m=\frac{F}{a}=\frac{75}{1.5}=50 kg$

3 0

2 years ago

An object, initially at rest, is subject to an acceleration of 34 m/s^2. How long will it take for that object to reach 3400m ?

Norma-Jean [14]

Vf^2 = Vi^2 + 2ad
a= 34 m/s^2
Vi = 0 m/s
d = 3400m

Vf = 480.83 m/s

a=v/t
t=v/a
t=480.83/34
t=14.142 s

6 0

2 years ago

The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at

denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m _____ eqn (1)

FOR HYDROGEN:

v = √(3KT)/m = 2000 m/s _____ eqn (2)

FOR OXYGEN:

velocity of oxygen = √(3KT)/(mass of oxygen)

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

velocity of oxygen = (1/4)(2000 m/s) = 500 m/s

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

K.E of Oxygen = 4000 J

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

v = 2149.24 m/s

8 0

3 years ago

If the current through a 20-ω resistor is 8.0 a , how much energy is dissipated by the resistor in 1.0 h ?

fiasKO [112]

1 hour = 3600 seconds.
Energy dissipated = I²Rt = 8²×20×3600 = 4608000 J

4 0

2 years ago