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vampirchik [111]

2 years ago

15

A boy is exerting a force of 70 N at a 50-degree angle on a lawn mower. He is accelerating at 1.8 m/s2. Round the answers to the

nearest whole number.
What is the mass of the lawn mower?
kg
What is the normal force exerted on the lawn mower?
N

1 answer:

charle [14.2K]2 years ago

8 0

Answer:

are you in middle school

Explanation:

need more evendice

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A bowling ball because it is heavier and it has more air force going against it<span />

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How much tension must a rope withstand if it is used to accelerate a 6526 kg car vertically upwards at 8.9 m/s^2?

ExtremeBDS [4]

Answer:

The value is $T = 122036.2 \ N$

Explanation:

From the question we are told that

The mass of the car is $m = 6526 \ kg$

The acceleration is $a= 8.9 \ m/s^2$

Generally the net force applied on the rope is mathematically represented as

$F_{net} = T - W$

Here W is the weight of the car which is evaluated as

$W = m * g$

=> $W = 6526 * 9.8$

=> $W = 63954.8 \ N$

Generally the net force can also be mathematically represented as

$F = m * a$

So

$m * a = T - 63954.8$

=> $6526 * 8.9 = T - 63954.8$

=> $T = 122036.2 \ N$

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You are trying to overhear a most interesting conversation, but from your distance of 10.0 m , it sounds like only an average wh

Alexandra [31]

Answer:

r₂=0.1 m

Explanation:

Given that

r₁= 10 m , β₁ = 20 dB

At r₂ ,β₂= 60 dB

As we know that intensity level of sound given as

$\beta =10\ log\dfrac{I}{10^{-12}}$

$\beta _1=10\ log\dfrac{I_1}{10^{-12}}$

$20=10\ log\dfrac{I_1}{10^{-12}}$

10² x 10⁻¹² = I₁

I₁=10⁻¹⁰ W/m²

$\beta _2=10\ log\dfrac{I_2}{10^{-12}}$

$60=10\ log\dfrac{I_1}{10^{-12}}$

10⁶ x 10⁻¹² = I₂

I₂ = 10⁻⁶ W/m²

I₁=10⁻¹⁰ W/m²

P = I A

P=Power ,I =Intensity ,A=Area

$\dfrac{I_1}{I_2}=\dfrac{r^2_2}{r^2_1}$

$\dfrac{10^{-10}}{10^{-6}}=\dfrac{r^2_2}{10^2}$

r₂=0.1 m

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2 years ago

Suppose a yo-yo has a center shaft that has a 0.230 cm radius and that its string is being pulled.

Fofino [41]

Answer:

Part a)

$\alpha = 782.6 rad/s^2$

Part B)

$\omega = 587 rad/s$

Part c)

$a_t = 24.3 m/s^2$

Explanation:

Part a)

As we know that

$a = R \alpha$

so we will have

$a = 1.80 m/s^2$

$R = 0.230 cm$

$\alpha = \frac{a}{R}$

$\alpha = \frac{1.80}{0.230 \times 10^{-2}}$

$\alpha = 782.6 rad/s^2$

Part B)

Angular speed of the yo-yo

$\omega = \alpha t$

so we have

$\omega = 782.6 \times 0.750$

$\omega = 587 rad/s$

Part c)

Tangential acceleration is given as

$a_t = R \alpha$

$a_t = (3.10 \times 10^{-2})(782.6)$

$a_t = 24.3 m/s^2$

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3 years ago