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2 years ago

Help me with the question in the image I provided

Physics

1 answer:

inna [77]2 years ago

8 0

Answer: 10 s, 30 m/s , 150 m

Explanation:

Given

The speed of motorcyclist is $u_1=15\ m/s$

The initial speed of a police motorcycle is $u_2=0\ m/s$

acceleration of police motorcycle is $a=3\ m/s^2$

Police will catch the motorcyclist when they traveled equal distances

distance traveled by motorcyclist in time t is

$\Rightarrow s_1=15\times t$

Distance traveled by Police in time t is

$\Rightarrow s_2=u_2t+\frac{1}{2}at^2\\\Rightarrow s_2=0+0.5\times 3\times t^2$

put $s_1=s_2$

$\Rightarrow 15t=0.5\times 3\times t^2\\\\\Rightarrow t(1.5t-15)=0\\\\\Rightarrow t=\dfrac{15}{1.5}=10\ s$

Police officer's speed at that time is

$\Rightarrow v_2=u_2+at\\\Rightarrow v_2=0+3\times 10=30\ m/s$

Distance traveled by each vehicle is

$\Rightarrow s_1=s_2=15\times t=15\times 10=150\ m$

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What compound makes wood

Luda [366]

Answer:lignin

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2 years ago

A 75.0-kg man standing on a bathroom scale in an elevator. Calculate the scale in N, reading if the elevator moves upward at a c

Salsk061 [2.6K]

The scale in N, reading if the elevator moves upward at a constant speed of 1.5 m/s^2 is 862.5 N.

weight of man = 75kg

speed of elevator, a = 1.5 $m/ s^{2}$

$F - w = ma \\$

$F = w + ma$

$F = m ( a +g )$

$F = 75 ( 1.5 + 10 ) \\$

$F = 75 ( 11.5 )$

$F = 862.5 N$

So, the scale reading in the elevator is greater than his 862.5 N weight. This indicates that the person is being propelled upward by the scale, which it must do in order to do so, with a force larger than his weight. According to what you experience in quickly accelerating or slowly moving elevators, it is obvious that the faster the elevator acceleration, the greater the scale reading.

Speed can be defines as the pace at which the position of an object changes in any direction. Since speed simply has a direction and no magnitude, it is a scalar quantity.

Learn more about speed here:-

brainly.com/question/19127881

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8 0

1 year ago

Suppose a 4.0-kg projectile is launched vertically with a speed of 8.0 m/s. What is the maximum height the projectile reaches?

eduard

Answer:

h = 3.3 m (Look at the explanation below, please)

Explanation:

This question has to do with kinetic and potential energy. At the beginning (time of launch), there is no potential energy- we assume it starts from the ground. There, is, however, kinetic energy

Kinetic energy = $\frac{1}{2}$ m $v^{2}$

Plug in the numbers = $\frac{1}{2}$ (4.0)( $8^{2}$ )

Solve = 2(64) = 128 J

Now, since we know that the mechanical energy of a system always remains constant in the absence of outside forces (there is no outside force here), we can deduce that the kinetic energy at the bottom is equal to the potential energy at the top. Look at the diagram I have attached.

Potential energy = mgh = (4.0)(9.8)(h) = 39.2(h)

Kinetic energy = Potential Energy

128 J = 39.2h

h = 3.26 m

h= 3.3 m (because of significant figures)

7 0

2 years ago

A 40.0 kg beam is attached to a wall with a hi.nge and its far end is supported by a cable. The angle between the beam and the c

GalinKa [24]

The magnitude of the force that the beam exerts on the hi.nge will be,261.12N.

To find the answer, we need to know about the tension.

<h3>How to find the magnitude of the force that the beam exerts on the hi.nge?</h3>

Let's draw the free body diagram of the system using the given data.
From the diagram, we have to find the magnitude of the force that the beam exerts on the hi.nge.
For that, it is given that the horizontal component of force is equal to the 86.62N, which is same as that of the horizontal component of normal reaction that exerts by the beam on the hi.nge.

$N_x=86.62N$

We have to find the vertical component of normal reaction that exerts by the beam on the hi.nge. For this, we have to equate the total force in the vertical direction.

$N_y=F_V=mg-Tsin59\\$