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viktelen [127]
3 years ago
10

A mouse jumps horizontally from a box of height 0.25m.  If the mouse jumps with a speed of 2.1 m/s, how far from the box does th

e mouse land?
Physics
2 answers:
Nookie1986 [14]3 years ago
7 0

Time to fall 0.25 meter vertically from rest:

D = (1/2) (g) (t²)

0.25m = (1/2) (9.8 m/s²)(t²)

t² = (0.25m) / (4.9 m/s²)

t² = 0.051 sec²

t = 0.226 second

In that amount of time, the mouse sails out

(2.1 m/s) (0.226 sec) = <em>0.474 meter</em>

horizontally.


Sladkaya [172]3 years ago
3 0

The mouse would land 0.47 m away from the box.

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You are on a 4 m high ladder and throw a ball upwards at 12 m/s. It lands on the ground below the ladder.
Gelneren [198K]

a) 2.75 s

The vertical position of the ball at time t is given by the equation

y= h+ut-\frac{1}{2}gt^2

where

h = 4 m is the initial height of the ball

u = 12 m/s is the initial velocity of the ball (upward)

g = 9.8 m/s^2 is the acceleration of gravity (downward)

We can find the time t at which the ball reaches the ground by substituting y=0 into the equation:

0 = 4 + 12t - 4.9 t^2

This is a second-order equation. By solving it for t, we find:

t = -0.30 s

t = 2.75 s

The first solution is negative, so we discard it; the second solution, t = 2.75 s, is the one we are looking for.

b) -15.0 m/s (downward)

The final velocity of the ball can be calculated by using the equation:

v=u-gt

where

u = 12 m/s is the initial (upward) velocity

g = 9.8 m/s^2 is the acceleration of gravity (downward)

t is the time

By subsisuting t = 2.75 s, we find the velocity of the ball as it reaches the ground:

v=12 -(9.8)(2.75)=-15.0 m/s

And the negative sign means the direction is downward.

7 0
3 years ago
A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calcu
Alekssandra [29.7K]

Complete question:

A wire 2.80 m in length carries a current of 5.60 A in a region where a uniform magnetic field has a magnitude of 0.300 T. Calculate the magnitude of the magnetic force on the wire assuming the following angles between the magnetic field and the current.

a) 60 ⁰

b) 90 ⁰

c) 120 ⁰

Answer:

(a) When the angle, θ = 60 ⁰,  force = 4.07 N

(b) When the angle, θ = 90 ⁰,  force = 4.7 N

(c) When the angle, θ = 120 ⁰,  force = 4.07 N

Explanation:

Given;

length of the wire, L = 2.8 m

current carried by the wire, I = 5.6 A

magnitude of the magnetic force, F = 0.3 T

The magnitude of the magnetic force is calculated as follows;

F = BIl \ sin(\theta)

(a) When the angle, θ = 60 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(60)\\\\F = 4.07 \ N

(b) When the angle, θ = 90 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(90)\\\\F = 4.7 \ N

(c) When the angle, θ = 120 ⁰

F = BIl \ sin(\theta)\\\\F = 0.3 \times 5.6 \times 2.8 \times sin(120)\\\\F = 4.07 \ N

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3 years ago
Plzz answer the question.
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Ur answer is 3 and i'm sure of it 
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3 years ago
The electric field is measured for points at distances r from the center of a uniformly charged insulating sphere that has volum
RoseWind [281]

For the electric field is measured for points at distances r,Electric field  is mathematically given as

E=19.9*10^{-5}c/m^3

<h3>What is the Electric field?</h3>

Generally, the equation for the  electric field uniform charged is mathematically given as

E=\frac{lr}{3e0}

Therefore

E=lr/3E0=3Ee0/r

Therefore

E=3*6*10^4*8.85*10^{-12}

E=19.9*10^{-5}c/m^3

In conclusion, E=19.9*10^{-5}c/m^3

E=19.9*10^{-5}c/m^3

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In a hydraulic lift, the radius of the small and large pistons are
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Answer:

the answer is 52.15cm×35.5

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