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4 years ago

A block is pulled

Physics

1 answer:

Natasha_Volkova [10]4 years ago

6 0

Answer:

if your teacher taught you im pretty sure you add the numbers together

Explanation:im wrong dont listen to me

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Plzzzz help me this is due today. only need help with these two questions and the element is oxygen.

Tresset [83]

Answer: thanks for the point

Explanation:

3 0

3 years ago

1) A net force of 75.5 N is applied horizontally to slide a 225 kg crate across the floor.

dimulka [17.4K]

Answer:

The acceleration of the crate is $0.3356\,\frac{m}{s^2}$

Explanation:

Recall the formula that relates force,mass and acceleration from newton's second law;

$F=m\,a$

Then in our case, we know the force applied and we know the mass of the crate, so we can solve for the acceleration as shown below:

$F=m\,a\\75.5\,N=225\,\,kg\,\,a\\a=\frac{75.5}{225} \,\frac{m}{s^2} \\a=0.3356\,\,\frac{m}{s^2}$

7 0

3 years ago

do u ever think that how are u living cause we could not even be here and God but made us but had did it all started I believe i

Katena32 [7]

Answer:

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Explanation:

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5 0

3 years ago

A person pushes horizontally with a force of 220. N on a 61.0 kg crate to move it across a level floor. The coefficient of kinet

Ede4ka [16]

Answer:

(a) 161.57 N

(b) 0.958 m/s^2

Explanation:

Force applied, F = 220 N

mass of crate, m = 61 kg

μ = 0.27

(a) The magnitude of the frictional force,

f = μ N

where, N is the normal reaction

N = m x g = 61 x 9.81 = 598.41 N

So, the frictional force, f = 0.27 x 598.41

f = 161.57 N

(b) Let a be the acceleration of the crate.

Fnet = F - f = 220 - 161.57

Fnet = 58.43 N

According to newton's second law

Fnet = mass x acceleration

58.43 = 61 x a

a = 0.958 m/s^2

Thus, the acceleration of the crate is 0.958 m/s^2.

7 0

3 years ago

Suppose that the sound level of a conversation is initially at an angry 71 dB and then drops to a soothing 54 dB. Assuming that

Goryan [66]

Answer:

Explanation:

In the decibel scale , intensity of sound changes logarithmically as follows

$10log\frac{I}{I_0} =$ Value in decibel scale , the value of I₀ = 10⁻¹² W /m².

Putting the values

$10log\frac{I}{10^{-12}} = 71$

$log\frac{I}{10^{-12}} = 7.1$

$\frac{I}{10^{-12}} = 10^{7.1}$

$I= 10^{-4.9}$ W/m²

Similarly for 54 dB sound intensity can be given as follows

I = 10⁻¹² x $10^{5.4}$

$I= 10^{-6.6 }$ W / m²

For intensity of sound the relation is as follows

I = 2π²υ²A²ρc where υ is frequency , A is amplitude , ρ is density of air and c is velocity of sound .

Putting the given values for 71 dB

$I= 10^{-4.9}$ = 2π² x 504²xA²x 1.21 x 346

A² = 60.03 x 10⁻¹⁶

A = 7.74 x 10⁻⁸ m

For 54 dB sound

$10^{-6.6}$ = 2π² x 504²xA²x 1.21 x 346

A² = 1.1978 x 10⁻¹⁶

A = 1.1 x 10⁻⁸ m

6 0

4 years ago