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2 years ago

A projectile is shot on level ground with a

Physics

1 answer:

erastova [34]2 years ago

5 0

Answer:

time rising = 34 / 9.8 = 3.47 sec

total time in air = 2 * 3.47 sec = 6.94 sec

(time rising must equal time falling)

R = 17 m/s * 6.94 s = 118 m

Can also use range formula

R = v^2 sin (2 theta) / g

tan theta = 34 / 17 = 2

theta = 63.4 deg

2 theta = 126.9 deg

sin 126.9 = .8

v^2 = 17^2 + 34^2 = 1445 m^2/s^2

R = 1445 * .8 / 9.8 = 118 m agreeing with answer found above

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A man weighs himself twice in an elevator. When the elevator is at rest, he weighs 824 N; when the elevator starts moving upward

kifflom [539]

Answer: c. 1.3 m/s^2

Explanation:

When he is at rest, is weight can be calculated as:

W = g*m

where:

m = mass of the man

g = gravitational acceleration = 9.8m/s^2

We know that at rest his weight is W = 824N, then we have:

824N = m*9.8m/s^2

824N/(9.8m/s^2) = m = 84.1 kg

Now, when the elevators moves up with an acceleration a, the acceleration that the man inside fells down is g + a.

Then the new weight is calculated as:

W = m*(g + a)

and we know that in this case:

W = 932N

g = 9.8m/s^2

m = 84.1 kg

Then we can find the value of a if we solve:

932N = 84.1kg*(9.8m/s^2 + a)

932N/84.1kg = 11.1 m/s^2 = 9.8m/s^2 + a

11.1 m/s^2 - 9.8m/s^2 = a = 1.3 m/s^2

The correct option is C

3 0

2 years ago

What is the term to describe the rate of flow of electricity?

SVEN [57.7K]

The rate of flow of electric CHARGE past any point is described in the unit of electric CURRENT ... the Ampere.

7 0

2 years ago

To analyze the motion of a body that is traveling along a curved path, to determine the body's acceleration, velocity, and posit

DiKsa [7]

To solve this problem we will apply the kinematic equations of linear motion and centripetal motion. For this purpose we will be guided by the definitions of centripetal acceleration to relate it to the tangential velocity. With these equations we will also relate the linear velocity for which we will find the points determined by the statement. Our values are given as

$R = 350ft$

$a_t = 1.1ft/s^2$

PART A )

$a_c = \frac{V^2}{R}$

$a_c = \frac{V^2}{350}$

Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is $5.25ft/s^2$

$a = \sqrt{a_t^2+a_r^2}$

$5.25 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}$

$27.5625 = 1.21 + \frac{v^4}{122500}$

$v=42.3877ft/s$

Now calculate the angular velocity of the motorcycle

$v = r\omega$

$42.3877 = 350\omega$

$\omega = 0.1211rad/s$

Calculate the angular acceleration of the motorcycle

$a_t = r\alpha$

$1.1 = 350\alpha$

$\alpha = 3.1428*10^{-3}rad/s^2$

Calculate the time needed by the motorcycle to reach an acceleration of

$5.25ft/s^2$

$\omega = \alpha t$

$0.1211 = 3.1428*10^{-3}t$

$t = 38.53s$

PART B) Calculate the velocity of the motorcycle when the net acceleration of the motorcycle is $6.75ft/s^2$

$a = \sqrt{a_t^2+a_r^2}$

$6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}$

$45.5625 = 1.21 + \frac{v^4}{122500}$

$v=48.2796ft/s$

PART C)

Calculate the radial acceleration of the motorcycle when the velocity of the motorcycle is $21.5ft/s$

$a_r = \frac{v^2}{R}$

$a_r = \frac{21.5^2}{350}$

$a_r =1.3207ft/s^2$

Calculate the net acceleration of the motorcycle when the velocity of the motorcycle is $21.5ft/s$

$a = \sqrt{a_t^2+a_r^2}$

$a = \sqrt{(1.1)^2+(1.3207)^2}$

$a = 1.7187ft/s^2$

PART D) Calculate the maximum constant speed of the motorcycle when the maximum acceleration of the motorcycle is $6.75ft/s^2$

$a = \sqrt{a_t^2+a_r^2}$

$6.75 = \sqrt{(1.1)^2+(\frac{v^2}{350})^2}$

$45.5625 = 1.21 + \frac{v^4}{122500}$

$v=48.2796ft/s$

3 0

2 years ago

1500 kg wrecking ball traveling at a speed of 3.5 m/s hits a wall that does not crumble but is pushed back 75 cm. If the wreckin

Rudiy27

Answer:

The size of the force that pushes the wall is 12,250 N.

Explanation:

Given;

mass of the wrecking ball, m = 1500 kg

speed of the wrecking ball, v = 3.5 m/s

distance the ball moved the wall, d = 75 cm = 0.75 m

Apply the principle of work-energy theorem;

Kinetic energy of the wrecking ball = work done by the ball on the wall

¹/₂mv² = F x d

where;

F is the size of the force that pushes the wall

¹/₂mv² = F x d

¹/₂ x 1500 x 3.5² = F x 0.75

9187.5 = 0.75F

F = 9187.5 / 0.75

F = 12,250 N

Therefore, the size of the force that pushes the wall is 12,250 N.

7 0

2 years ago

The display of the aurora and the reflection of radio waves back to earth result from the _____.

Tasya [4]

The layer of electrically charged molecules and atoms which spans 40-250 miles above ground called ionosphere causes the display of the aurora and the reflection of radio waves back to earth.

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3 years ago

Read 2 more answers