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2 years ago

A grasshopper jumps at a 65.0 degree angle at 5.42m/s. At what time does it reach its maximum height?

Physics

2 answers:

dimulka [17.4K]2 years ago

8 0

Answer:

A grasshopper reached the maximum height 1.23 m.

Explanation:

Given that,

Velocity = 5.42 m/s

Angle = 65.0°

We need to calculate the maximum height

Using formula of maximum height

$y_{max}=\dfrac{v^2\sin^2\theta}{2g}$

Where, v = 5.42 m/s

g = acceleration due to gravity

$\theta$ =Angle

Put the value into the formula

$y_{max}=\dfrac{5.42^2\times\sin^2(65.0)}{2\times9.8}$

$y_{max}=1.23\ m$

Hence, A grasshopper reached the maximum height 1.23 m.

crimeas [40]2 years ago

6 0

When the grasshoppers vertical velocity is exactly zero.
v = -g•t + v0.
v: vertical part of velocity. Is zero at maximum height.
g: 9.81
t: time you are looking for
v0: initial vertical velocity
Find the vertical part of the initial velocity, by using the angle at which the grasshopper jumps.

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In a circus act, an acrobat rebounds upward from the surface of a trampoline at the exact moment that another acrobat, perched 9

slega [8]

Answer:

1.6 secs

Explanation:

In a circus act, an acrobat upwards from the surface of a trampoline

At that same moment another acrobat perched 9.0m above him

A ball is released from rest

While still in motion the acrobat catches the ball

He left the ball with a trampoline of 5.6m/s

Since the ball is falling downwards from a distance then acceleration will be negative

a= -9.8

s= d

s= 1/2at^2

= 1/2 × (-9.8)t^2

= 0.5× (-9.8)t^2

d = -4.9t^2

Therefore the time the acrobat stays in the air before catching the ball can be calculated as follows

9 - 4.9t^2= 5.6t + 1/2(-9.8)t^2

9 - 4.9t^2= 5.6t + (-4.9)t^2

9 - 4.9t^2= 5.6t - 4.9t^2

9= 5.6t

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2 years ago

What is the length of the x-component of the vector shown below?

jeka94

Answer:

Explanation:

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2 years ago

Mars has two moons, Phobos and Deimos. Phobos orbits Mars at a distance of 9380 km from Mars's center, while Deimos orbits at 23

Sloan [31]

Answer:

The ratio is $\frac{T_1}{T_2} = 3.965$

Explanation:

From the question we are told that

The radius of Phobos orbit is R_2 = 9380 km

The radius of Deimos orbit is $R_1 = 23500 \ km$

Generally from Kepler's third law

$T^2 = \frac{ 4 * \pi^2 * R^3}{G * M }$

Here M is the mass of Mars which is constant

G is the gravitational constant

So we see that $\frac{ 4 * \pi^2 }{G * M } = constant$

$T^2 = R^3 * constant$

=> $[\frac{T_1}{T_2} ]^2 = [\frac{R_1}{R_2} ]^3$

Here $T_1$ is the period of Deimos

and $T_1$ is the period of Phobos

$[\frac{T_1}{T_2} ] = [\frac{R_1}{R_2} ]^{\frac{3}{2}}$

=> $\frac{T_1}{T_2} = [\frac{23500 }{9380} ]^{\frac{3}{2}}]$

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2 years ago

In the late 19th century, great interest was directed toward the study of electrical discharges in gases and the nature of so-ca

gizmo_the_mogwai [7]

Answer:

A) He finds the same value of q / m for different materials , B) y = ½ (q / m) E L² / v₀ₓ² , C) v = E / B , D) B = 2.13 10⁻⁶ T, E) For the first part I have two off-center points., For the second part I can center one point but the other is off center

Therefore the third statement is correct

Explanation:

Part A

Thomson's experiments are the first proof that the atoms that until now were considered indivisible were constituted by different elements, in these experiments Thomson himself the ratio q / m of several cathodes and always found the same value, which allowed to establish that In atoms there are two types of particles, some of which are mobile and others are still.

When examining his statements the correct one is: He finds the same value of q / m for different materials

Part B

For this part let's use Newton's second law

F = ma

q E = m a

a = q / m E

We use the kinematic relationship

y = voy t - ½ to t²

x = v₀ₓ t

The initial vertical velocity of electrons is zero

y = ½ a (x / v₀ₓ)²

We replace

y = ½ (q / m) E L² / v₀ₓ²

Part C

If there is no deflection, the electric and magnetic forces are the same and in the opposite direction

Fm = Fe

q v B = q E

v = E / B

Part D

We replace

y = ½ (q / m) E L² / (E / B)²

y = ½ (q / m) L² B² / E

If we do not want any deflection the magnetic field has to return the electrons the amount that they lower y = -4.12 cm

-4.12 10⁻² = ½ q / m 0.12² B² / 1.1 10³

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B = √ 0.259 1.758 10¹¹ = √ 4.55 10⁻¹²

B = 2.13 10⁻⁶ T

Part E

As the charge that the two particles is different

For the first part I have two off-center points.

For the second part I can center one point but the other is off center

Therefore the third statement is correct

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2 years ago