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3 years ago

PLEASE HELPPP!!!!! I need help immediately, some answer this question

Physics

1 answer:

Alex17521 [72]3 years ago

5 0

Answer:

A)It would stay stationery as the resultant force is 0.

B)It would move towards the right as the resultant force is 300N due right.

C)It would move left as the resultant force is 300N due left.

Explanation:

Please make it brainiest answer

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4. A 1,000 kg truck moving at 10 m/s runs into a concrete wall. It takes 0.5 seconds for the truck to conipietery

Maru [420]

Answer:

$\large \boxed{\text{h. 20 000 N}}$

Explanation:

Force is the change in momentum over time

F = Δp/Δt

1. Calculate the change in momentum

p₁ = mv₁ = 1000 kg × 10 m/s = 10 000 kg·m·s⁻¹

p₂ = 0

Δp = p₂ - p₁= (0 - 10 000) kg·m·s⁻¹ = -10 000 kg·m·s⁻¹

2. Calculate the force

$\begin{array}{rcl}F & = & \dfrac{\Delta p}{\Delta t}\\\\& = & \dfrac{-10 000 \text{ kg$\cdot$m$\cdot$ s}^{-1}}{\text{ 0.5 s}}\\\\& = & \textbf{-20 000 N}\\\end{array}\\\text{The negative sign shows that the force is exerted opposite to the direction of motion.}\\\text{The magnitude of the force is $\large \boxed{\textbf{20 000 N}}$}$

3 0

3 years ago

Read 2 more answers

Please do all of i will give you brainlest and thanks to best answer plz do it right

Sergeu [11.5K]

Answer:

winter solstice i think

Explanation:

8 0

3 years ago

Read 2 more answers

in a tall building, a certain window is 500 meters above the ground.A tennis ball is catapulted horizontally out of the window w

expeople1 [14]

The tennis ball lands at a point 40.4 m from the base of the building.

The tennis ball is projected with a horizontal velocity u from a window, which is at a height y from the ground. The ball lands at a distance x from the base of the building. Let the ball take a time t to reach the ground. In the time t ,the ball falls a vertical distance y and also travel a horizontal distance x.

The initial vertical velocity of the ball is zero, since the ball is projected in the horizontal direction. The ball falls down under the action of gravitational force.

Thus, use the equation of motion,

$y=\frac{1}{2} gt^2$

rewrite the expression for t and calculate the value of t using 9.81 m/s²for g and 500 m for y.

$t=\sqrt{\frac{2y}{g} } \\ =\sqrt\frac{(2)(500m)}{9.81m/s^2} \\ =10.096 s$

The horizontal distance x is traveled using the constant velocity u since no force acts on the ball in the horizontal direction.

Therefore,

$x=ut$

Substitute 4 m/s for u and 10.096 s for t

$x=ut\\ =(4m/s)(10.096s)\\ =40.384m=40.4 m$

Thus, the ball lands at a point 40.4 m from the base of the building.

6 0

3 years ago

A body is accelerated continuously. What is the form of the graph?

Inga [223]

That depends on what quantity is graphed.

It also depends on what kind of acceleration is taking place ...
continuous change of speed or continuous change of direction.

-- If the graph shows speed vs time, and the acceleration is a change
in speed, then the graph is a connected series of straight-line pieces.
Each straight piece slopes up if speed is increasing, or down if speed
is decreasing.

-- If the graph shows speed vs time, and the acceleration is a change in
direction only, then the graph is a straight horizontal line, since speed is
constant.

-- If the graph shows direction vs time, and the acceleration is a change
in speed only, then the graph is a straight horizontal line, since direction
is constant.

-- If the graph shows direction vs time, and the acceleration is a change
in direction, then the graph is a connected series of pieces of line.
Each piece may be straight if the direction is changing at a constant rate,
or curved if the direction is changing at a rate which grows or shrinks.
Each piece may slope up if the angle that defines the direction is growing,
or may slope down if the angle that defines the direction is decreasing.

-- If the graph shows distance vs time, and the acceleration is a
change in speed, then the graph is a connected series of pieces
of curves. Each piece curves up if speed is increasing, or down if
speed is decreasing.

-- If the graph shows distance vs time, and the acceleration is a change
in direction only, then the graph is a straight line sloping up, since speed
is constant.

6 0

3 years ago

The Sun’s surface temperature is about 5800 K and its spectrum peaks at 5000 Å. An O-type star’s surface temperature may be 40,0

nirvana33 [79]

(a) $7.25\cdot 10^{-8}m$

Wien's displacement law is summarized by the equation

$\lambda = \frac{b}{T}$

where

$\lambda$ is the peak wavelength

$b=2.898\cdot 10^{-3} m \cdot K$ is Wien's displacement constant

T is the absolute temperature at the surface of the star

For an O-type star, we have

T = 40,000 K

Therefore, its peak wavelength is

$\lambda = \frac{2.898\cdot 10^{-3}}{40000}=7.25\cdot 10^{-8}m$

(b) Ultraviolet

We can answer this part by looking at the wavelength range of the different parts of the electromagnetic spectrum:

gamma rays

X-rays 1 nm - 1 pm

ultraviolet 380 nm - 1 nm

visible light 750 nm - 380 nm

infrared $25 \mu m - 750 nm$

microwaves 1 mm - $25 \mu m$

radio waves > 1 mm

The peak wavelength of this star is

$\lambda=7.25\cdot 10^{-8}m=72.5 nm$

Therefore, it falls in the ultraviolet region.

(c) No

The Keck telescopes is actually a system of 2 telescopes in the Keck Observatory, located in Mauna kea, Hawai.

The two telescopes, thanks to several instruments, are able to detect much of the electromagnetic radiation in the visible ligth and infrared parts of the spectrum. However, they are not able to detect light in the ultraviolet region: therefore, they cannot observe the star mentioned in the previous part of the problem.

7 0

3 years ago

PLEASE HELPPP!!!!! I need help immediately, some answer this question​

PLEASE HELPPP!!!!! I need help immediately, some answer this question