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3 years ago

5

Do air bags increase pressure on the body during an accident

1 answer:

victus00 [196]3 years ago

7 0

Yes because once the accident has happen the air bag will then push out

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The best scientific reason for a scientist to accept a specific theory is?

lions [1.4K]

Answer:

The theory is supported by all the available observations and data.

Explanation: The scientific community will accept a theory when a sufficient body of evidence supports it. This includes experiments that refute other potential theories. Experiments should also be carried out that attempt to disprove the theory but cannot.

It should not matter who proposed the theory or who supports it, and instead should entirely be based on the quality and abundance of data supporting it.

Hope this helps!

Pls Mark me brainliest

-ya girl Lizzie

7 0

2 years ago

Two objects are moving at equal speed along a level, frictionless surface. the second object has twice the mass of the first obj

lbvjy [14]

Answer:

They both rises to same height.

Explanation:

When an object is sliding up in friction less surface than according to conservation of energy its potential energy will be converted into kinetic energy.

$mgH=\frac{1}{2}mv^{2}\\ v=\sqrt{2gH}$

Here, m is the mass, v is the velocity, g is the acceleration due to gravity and H is the height.

Here the height is independent on the mass of an object and its only depend on velocity.

Now according to the question, two objects have same velocity but they have different masses.

Therefore, they rises to the same height because height will not change with mass.

8 0

3 years ago

Read 2 more answers

Samantha is checking the weather for her upcoming trip to Mexico City. The weather forecast predicts a high-pressure system for

Zanzabum

Calm, sunny days with wind moving away from the center.

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2 years ago

Read 2 more answers

Larry drops a 5kg ball off of a building. The ball hits the ground 4.7s later. How tall is the building?

musickatia [10]

Explanation:

Given:

v₀ = 0 m/s

a = 9.8 m/s²

t = 4.7 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (4.7 s) + ½ (9.8 m/s²) (4.7 s)²

Δy ≈ 110 m

8 0

3 years ago

Consider a father pushing a child on a playground merry-go-round. The system has a moment of inertia of 84.4 kg.m^2. The father

Sophie [7]

Answer:

Explanation:

Given that:

the initial angular velocity $\omega_o = 0$

angular acceleration $\alpha$ = 4.44 rad/s²

Using the formula:

$\omega = \omega_o+ \alpha t$

Making t the subject of the formula:

$t= \dfrac{\omega- \omega_o}{ \alpha }$

where;

$\omega = 1.53 \ rad/s^2$

∴

$t= \dfrac{1.53-0}{4.44 }$

t = 0.345 s

b)

Using the formula:

$\omega ^2 = \omega _o^2 + 2 \alpha \theta$

here;

$\theta$ = angular displacement

∴

$\theta = \dfrac{\omega^2 - \omega_o^2}{2 \alpha }$

$\theta = \dfrac{(1.53)^2 -0^2}{2 (4.44) }$

$\theta =0.264 \ rad$

Recall that:

2π rad = 1 revolution

Then;

0.264 rad = (x) revolution

$x = \dfrac{0.264 \times 1}{2 \pi}$

x = 0.042 revolutions

c)

Here; force = 270 N

radius = 1.20 m

The torque = F * r

$\tau = 270 \times 1.20 \\ \\ \tau = 324 \ Nm$

However;

From the moment of inertia;

$Torque( \tau) = I \alpha \\ \\ Since( I \alpha) = 324 \ Nm. \\ \\ Then; \\ \\ \alpha= \dfrac{324}{I}$

given that;

I = 84.4 kg.m²

$\alpha= \dfrac{324}{84.4} \\ \\ \alpha=3.84 \ rad/s^2$

For re-tardation; $\alpha=-3.84 \ rad/s^2$

Using the equation

$t= \dfrac{\omega- \omega_o}{ \alpha }$

$t= \dfrac{0-1.53}{ -3.84 }$

$t= \dfrac{1.53}{ 3.84 }$

t = 0.398s

The required time it takes= 0.398s

5 0

2 years ago