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BaLLatris [955]

3 years ago

13

A 44 kg block lies on a horizontal frictionless surface. A horizontal force of 110 N is applied to the block. (a) Determine the

acceleration of the block. (b) Calculate the distance the block will travel is the force is applied for 8 s. (c) Calculate the speed of the block after the force has been applied for 8 s.

1 answer:

Oksana_A [137]3 years ago

8 0

44k jkjk know the answer

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A stone tumbles into a mine shaft and strikes bottom after falling for 4.2 second how deep is the mine shaft

rjkz [21]

$4.2*9.8\\41.16$

41.16 meters

5 0

3 years ago

jet is flying at 500 mph east relative to the ground. A Cessna is flying at 150 mph 60° north of west relative to the ground. Wh

Greeley [361]

Answer:

C. 590 mph

$\vert v_{cj}\vert=589.49\ mph$

Explanation:

Given:

velocity of jet, $v_j=500\ mph$
direction of velocity of jet, east relative to the ground
velocity of Cessna, $v_c=150\ mph$
direction of velocity of Cessna, 60° north of west

Taking the x-axis alignment towards east and hence we have the velocity vector of the jet as reference.

Refer the attached schematic.

So,

$\vec v_j=500\ \hat i\ mph$

&

$\vec v_c=150\times (\cos120\ \hat i+\sin120\ \hat j)$

$\vec v_c=-75\ \hat i+75\sqrt{3}\ \hat j\ mph$

Now the vector of relative velocity of Cessna with respect to jet:

$\vec v_{cj}=\vec v_j-\vec v_c$

$\vec v_{cj}=500\ \hat i-(-75\ \hat i+75\sqrt{3}\ \hat j )$

$\vec v_{cj}=575\ \hat i-75\sqrt{3}\ \hat j\ mph$

Now the magnitude of this velocity:

$\vert v_{cj}\vert=\sqrt{(575)^2+(75\sqrt{3} )^2}$

$\vert v_{cj}\vert=589.49\ mph$ is the relative velocity of Cessna with respect to the jet.

8 0

3 years ago

How does temperature increase?

vovikov84 [41]

Increasing the temperature increases reaction rates because of the disproportionately increase in the number of high energy collisions.
It's only these collisions (possessing at least the activation energy for the reaction)
which results in a reaction!

6 0

2 years ago

Read 2 more answers

Write the equation expressing the relationship "y varies directly as x." use k as the constant of proportionality.

Alexeev081 [22]

The equation expressing the statement "y varies directly as x" is $y=kx.$

Explanation

The statement that $y$ varies directly as $x$ is analogous to saying that the ratio of $y$ to $x$ is constant. In other words, when $x$ increases, $y$ likewise increases and that when $x$ decreases, $y$ decreases proportionally.

Mathematically, we express the relationship that the ration of $y$ is to $x$ is constant is expressed as; $\frac{y}{x} =k$ where $k$ is the constant of proportionality.

We can then solve the relationship for $y$ to determine the correct form of the relationship as shown below,

$\frac{y}{x} =k\\\rightarrow y=kx$

6 0

2 years ago

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A radio have a wavelength of 0.3m and travels at a speed of 300,000,000 m/s. What is the frequency of this wave?

Ilya [14]

The frequency of the wave is $1\cdot 10^9 Hz$

Explanation:

The frequency, the wavelength and the speed of a wave are related by the following equation:

$c=f \lambda$

where

c is the speed of the wave

f is the frequency

$\lambda$ is the wavelength

For the radio wave in this problem,

$\lambda = 0.3 m$

$c=300,000,000 m/s = 3\cdot 10^8 m/s$

Therefore, the frequency is:

$f=\frac{c}{\lambda}=\frac{3\cdot 10^8}{0.3}=1\cdot 10^9 Hz$

Learn more about waves here:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

4 0

2 years ago