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4 years ago

1. If a 25 kg car accelerates at a speed of 100m/s2,2 what will the force of the car be?

Physics

1 answer:

algol134 years ago

4 0

Answer:

F = 2500N

Explanation:

F = ma

You're given m=25kg and a=100m/s²

Plug it in

F = (25)(100)

F = 2500N

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A 4.00 kg block is suspended from a spring with k 500 N/m. A 50.0 g bullet is fired into the block from directly below with a sp

Yanka [14]

Answer:

a. A = 0.1656 m

b. % E = 1.219

Explanation:

Given

mB = 4.0 kg , mb = 50.0 g = 0.05 kg , u₁ = 150 m/s , k = 500 N / m

To find the amplitude of the resulting SHM using conserver energy

ΔKe + ΔUg + ΔUs = 0

¹/₂ * m * v² - ¹/₂ * k * A² = 0

A = √ mB * vₓ² / k

vₓ = mb * u₁ / mb + mB

vₓ = 0.05 kg * 150 m / s / [0.050 + 4.0 ] kg = 1.8518

A = √ 4.0 kg * (1.852 m/s)² / (500 N / m)

A = 0.1656 m

The percentage of kinetic energy

%E = Es / Ek

Es = ¹/₂ * k * A² = 500 N / m * 0.1656²m = 13.72 N*0.5

Ek = ¹/₂ * mb * v² = 0.05 kg * 150² m/s = 1125 N

% E = 13.72 / 1125 = 0.01219 *100

% E = 1.219

6 0

3 years ago

A circular saw spins at 6000 rpm , and its electronic brake is supposed to stop it in less than 2 s. As a quality-control specia

Alona [7]

Answer:

a. yes

Explanation:

The initial speed of the circular saw is:

$\dot n_{o} = \frac{6000}{60} \,\frac{rev}{s}$

$\dot n_{o} = 100\,\frac{rev}{s}$

Deceleration rate needed to stop the circular saw is:

$\ddot n = -\frac{100\,\frac{rev}{s} }{2\,s}$

$\ddot n = - 50\,\frac{rev}{s^{2}}$

The number of turns associated with such deceleration rate is:

$\Delta n = \frac{\dot n^{2}}{2\cdot \ddot n}$

$\Delta n = \frac{\left(100\,\frac{rev}{s} \right)^{2}}{2\cdot \left(50\,\frac{rev}{s^{2}} \right)}$

$\Delta n = 100\,rev$

Since the measured number of revolutions is lesser than calculated number of revolution, the circular saw meets specifications.

8 0

3 years ago

In a machining operation that approximates orthogonal cutting, the cutting tool has a rake angle =100 . The chip thickness ratio

Rudik [331]

Answer:

The percentage of the total energy dissipated into shear plane is 89.46%.

Explanation:

Given that,

Rake angle = 10°

Thickness ratio= 0.5

Cutting Force = 400 N

Thrust force = 200 N

Speed =3 m/s

Suppose the shear force is 345.21 N.

We need to calculate the shear plane angle

Using formula shear angle

$\tan\phi=\dfrac{r\cos\alpha}{1-r\sin\alpha}$

Put the value in to the formula

$\tan\phi=\dfrac{0.5\cos10}{1-0.5\sin10}$

$\tan\phi=0.539$

$\phi=\tan^{-1}(0.539)$

$\phi=28.32^{\circ}$

We need to calculate the shear velocity

Using formula of shear velocity

$v_{2}=\dfrac{v\cos\alpha}{\cos(\phi-\alpha)}$

Put the value into the formula

$v_{2}=\dfrac{3\times\cos10}{\cos(28.32-10)}$

$v_{2}=3.11\ m/s$

We need to calculate the percentage of the total energy dissipated into shear plane

Using formula of energy dissipated

$\%d=\dfrac{P_{s}}{P}\times100$

$\%d=\dfrac{F_{s}\times v_{c}}{F_{c}\times v}\times100$

Put the value into the formula

$\%d=\dfrac{345.21\times3.11}{400\times3}\times100$

$d=89.46\%$

Hence, The percentage of the total energy dissipated into shear plane is 89.46%.

6 0

3 years ago

12- Calculate the power when a force of 60N moves an object over a distance of 0.6 km in 20

alexandr1967 [171]

Hi there!

To solve, we must begin by calculating the total WORK done on the object.

W = F · d (Force · displacement)

Plug in the given values. Remember to convert km to m:

1 km = 1000 m

0.6 km = 600 m

W = 60 · 600 = 36000 J

Now, we can solve for power:

P = W/t

Convert minutes to seconds:

1 min = 60 sec

20 min = 1200 sec

P = 36000/1200 = 30 W ⇒ Choice D.

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3 years ago

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3 years ago

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