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LuckyWell [14K]

2 years ago

7

An astronaut on a strange planet finds that she can jump a maximum horizontal distance of 15 m if her initial speed is 3 m/s. wh

at is the free-fall acceleration on the planet?

1 answer:

nata0808 [166]2 years ago

7 0

Answer:

R = V^2 sin 2 θ / g range formula

R is a maximum for θ = 45 and R = V^2 g

g = V^2 / R = 3^2 / 15 = .6 m/s^2

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Is a single property such as density sufficient to always correctly identify a substance?

bazaltina [42]

Not at all. Density on its own is not sufficient

7 0

3 years ago

What is the peak emf generated by rotating a 940-turn, 24 cm diameter coil in the Earth’s 5·10−5 T magnetic field, given th

elena-14-01-66 [18.8K]

Answer:

The peak emf generated by the coil is 2.67 V

Explanation:

Given;

number of turns, N = 940 turns

diameter, d = 24 cm = 0.24 m

magnetic field, B = 5 x 10⁻⁵ T

time, t = 5 ms = 5 x 10⁻³ s

peak emf, V₀ = ?

V₀ = NABω

Where;

N is the number of turns

A is the area

B is the magnetic field strength

ω is the angular velocity

V₀ = NABω and ω = 2πf = 2π/t

V₀ = NAB2π/t

A = πd²/4

V₀ = N x (πd²/4) x B x (2π/t)

V₀ = 940 x (π x 0.24²/4) x 5 x 10⁻⁵ x (2π/0.005)

V₀ = 940 x 0.04524 x 5 x 10⁻⁵ x 1256.8

V₀ = 2.6723 V = 2.67 V

The peak emf generated by the coil is 2.67 V

8 0

3 years ago

P6: An object of mass m sits on a spring of constant k in an elevator that is accelerating upwards with acceleration a. a) In te

tankabanditka [31]

Answer:

(a). The spring compressed is $\dfrac{ma+mg}{k}$ .

(b). The acceleration is 1.5 g.

Explanation:

Given that,

Acceleration = a

mass = m

spring constant = k

(a). We need to calculate the spring compressed

Using balance equation

$kx-mg=ma$

$x=\dfrac{ma+mg}{k}$ ....(I)

The spring compressed is $\dfrac{ma+mg}{k}$ .

(b). If the compression is 2.5 times larger than it is when the mass sits in a still elevator,

The compression is given by

$x=2.5\times x_{0}$

Here, acceleration is zero

So, $x=2.5\times\dfrac{mg}{k}$

We need to calculate the acceleration

Put the value of x in equation (I)

$2.5\times \dfrac{mg}{k}=\dfrac{ma+mg}{k}$

$2.5\times\dfrac{mg}{k}=\dfrac{m}{k}(a+g)$

$a=2.5g-g$

$a=1.5g$

Hence, (a). The spring compressed is $\dfrac{ma+mg}{k}$ .

(b). The acceleration is 1.5 g.

8 0

3 years ago

A block is projected up a frictionless inclined plane with initial speed v0 = 1.72 m/s. The angle of incline is θ = 44.8°. (a) H

Snowcat [4.5K]

Explanation:

Given

initial velocity(v_0)=1.72 m/s

$\theta =44.8{\circ}$

using $v^2-u^2=2as$

Where v=final velocity (Here v=0)

u=initial velocity(1.72 m/s)

a=acceleration $(gsin\theta )$

s=distance traveled

$0-(1.72)^2=2(-9.81\times sin(44.8))s$

s=0.214 m

(b)time taken to travel 0.214 m

v=u+at

$0=1.72-gsin(44.8)\times t$

$t=\frac{1.72}{9.8\times sin(44.8)}$

t=0.249 s

(c)Speed of the block at bottom

$v^2-u^2=2as$

Here u=0 as it started coming downward

$v^2=2\times gsin(44.8)\times 0.214$

$v=\sqrt{2.985}$

v=1.72 m/s

3 0

3 years ago

The lightest car in the world was built in London and had a mass of less than 10 kg. it's maximum speed was 25.0 km/h. Suppose t

Softa [21]

25km/h = 6.94 m/s
suvat
s=16
u=6.94
v=0
a=a
v^2=u^2+2as
(v^2-u^2)/2s = a =1.5ms^-2

6 0

3 years ago