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Novosadov [1.4K]

4 years ago

5

Emilie Sannom drives a rocket powered car to an initial speed of 228 mph(102 m/s) before attempting to make a jump of deadlizard

canyon on a ramp angled at 42 degrees. Assume that the rocket has burned out when she leaves the ramp. If the canyon is 420 m across and both sides are level, will she be able to make the jump? If so, how far from the edge will she land?

1 answer:

melamori03 [73]4 years ago

7 0

Answer:

Explanation:

Range= U²sin2theta/g

= 102²* sin (2*42)/9.8

= 1054.74m

Yes she will be abable to jump

The distance from the edge will be

(1054.74- 420)m

= 634.74m

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4 years ago

A roofer drops a nail that hits the ground traveling at 26 m/s. How fast was the nail traveling 1 second before it hits the grou

ELEN [110]

This problem can be solved using a kinematic equation. For this case, the following equation is useful:

v_final = v_initial + at

where,
v_final = final velocity of the nail
v_initial = initial velocity of the nail
a = acceleration due to gravity = 9.8 m/s^2
t = time

First, we determine the time it takes for the nail to hit the ground. We know that the initial velocity is 0 m/s since the nail was only dropped. It has a final velocity of 26 m/s. We substitute these values to the equation and solve for t:

26 = 0 + 9.8*t
t = 26/9.8 = 2.6531 s

The problem asks the velocity of the nail at t = 1 second. We then subtract 1 second from the total time 2.6531 with v_final as unknown.

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Thus, the nail was traveling at a speed of 16. 2004 m/s, 1 second before it hit the ground.

5 0

4 years ago

The image shows a model of the gravitational field around Earth. Which best describes the model?

GuDViN [60]

<u>Answer: The model is described by :</u>

The strength of the field increases as you get closer to Earth, as shown by the smaller spaces between the circles.

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According to universal gravitational law, the force acting in the universe between two objects is given by :

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It is clear that the gravitational force depends directly on the product of masses and inversely on the square of the distance between them.

The given image shows a model of the gravitational field around Earth. The arrows in the image show the direction of the force. Gravitational field depends on the position of objects.

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3 0

4 years ago

Read 2 more answers

A block is at rest on the incline shown in the gure. The coefficients of static and kinetic friction are 0.6 and 0.51, respec- t

Alekssandra [29.7K]

Normal reaction force on the block while it is at rest on the inclined plane is given as

$F_n = mgcos\theta$

here we know that

m = 46 kg

$\theta = 29^o$

now we will have

$F_n = 46*9.8*cos29 = 394.3 N$

now the limiting friction or maximum value of static friction on the block will be given as

$F_s = \mu_s * F_n$

$F_s = 0.6 * 394.3 = 236.56 N$

Above value is the maximum value of force at which block will not slide

Now the weight of the block which is parallel to inclined plane is given as

$F_{||} = mg sin\theta$

here we know that

$F_{||} = 46*9.8 sin29 = 218.55 N$

Now since the weight of the block here is less than the value of limiting friction force and also the block is at rest then the frictional force on the block is static friction and it will just counter balance the weight of the block along the inclined plane.

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3 years ago

Find the kinetic energy ofan electron that moves at half the speed of light to four significant digits.

Vlad1618 [11]

Answer:

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Explanation:

The electron has a mass of $9.1*10^{-31} kg$

The speed of light in a vacuum is a universal constant with the value 299 792 458 m / s (186 282,397 miles / s), although it is usually close to $3*10^{8} \frac{m}{s}$

Kinetic energy (K) is the energy associated with bodies that are in motion, depends on the mass and speed of the body and is calculated using the formula:

$K=\frac{1}{2} *m*v^{2}$ Equation(1)

K=kinetic energy (J)

m =mass of the body (kg)

v= speed of the body $(\frac{m}{s} )$

for this problem We replace in the equation (1)

$me=9.1*10^{-31} kg$ = electron mass

$ve=1.5*10^{8} \frac{m}{s}$ =Half the speed of light

=electron speed

We replace in the equation (1) :

$Ke=\frac{1}{2}*me*ve^{2}$

$Ke=\frac{1}{2} *9.1*10^{-31} *(1.5*10^{8} )^{2}$

$Ke=4.55*10^{-31} *(1.5)^{2} *(10^{8} )^{2}$

$Ke=10.24*10^{33} J$

The energy kinetic of the electron is $10.24*10^{33} Joules$

7 0

3 years ago