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2 years ago

Please help with this :(

Physics

1 answer:

Rainbow [258]2 years ago

3 0

Answer:

I don't really know but i have an idea

Explanation:

so acceleration means: increase in the rate or speed of something.

so just a head start.

your welcome!!

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during a baseball game you are running home and slide into home plate. However you come up short and you are tagged out. Which f

vovangra [49]

Answer:1 because

Explanation: it’s pointing to the earth and gravity

Pulls things down to earth

3 0

2 years ago

Calculate the acceleration of a mobile that at 4s is 32m from the origin, knowing that its initial speed is 10m / s.

ehidna [41]

Answer:

5.5 m/s^2

Explanation:

I believe this is the answer > using the formula a= v-v0/t

Hope this helps!

7 0

2 years ago

Read 2 more answers

A mass-spring system has k = 56.8 N/m and m = 0.46 kg.

andriy [413]

Answer:

A. $\omega=11.1121\ rad.s^{-1}$

B. $f=1.7685\ Hz$

C. $T=0.5654\ s$

Explanation:

Given:

spring constant, $k=56.8\ N.m^{-1}$

mass attached, $m=0.46\ kg$

for a spring-mass system the frequency is given as:

$\omega=\sqrt{\frac{k}{m} }$

$\omega=\sqrt{\frac{56.8}{0.46}}$

$\omega=11.1121\ rad.s^{-1}$

frequency is given as:

$f=\frac{\omega}{2\pi}$

$f=\frac{11.1121}{2\pi}$

$f=1.7685\ Hz$

Time period of a simple harmonic motion is given as:

$T=\frac{1}{f}$

$T=0.5654\ s$

7 0

2 years ago

In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. Ignore fr

spin [16.1K]

Answer:

a) $v_p=9.35m/s$

Explanation:

From the question we are told that:

Open cart of mass $M_o=50.0 kg$

Speed of cart $V=5.00m/s$

Mass of package $M_p=15.0kg$

Speed of package at end of chute $V_c=3.00m/s$

Angle of inclination $\angle =37$

Distance of chute from bottom of cart $d_x=4.00m$

Generally the equation for work energy theory is mathematically given by

$\frac{1}{2}mu^2+mgh=\frac{1}{2}mv_p^2$

Therefore

$\frac{1}{2}u^2+gh=\frac{1}{2}v_p^2$

$v_p=\sqrt{2(\frac{1}{2}u^2+gh)}$

$v_p=\sqrt{2(\frac{1}{2}v_c^2+gd_x)}$

$v_p=\sqrt{2(\frac{1}{2}(3)^2+(9.8)(4))}$

$v_p=9.35m/s$

4 0

2 years ago

An electric field of 1.27 kV/m and a magnetic field of 0.490 T act on a moving electron to produce no net force. If the fields a

lesantik [10]

Answer:

v = 2591.83 m/s

Explanation:

Given that,

The electric field is 1.27 kV/m and the magnetic field is 0.49 T. We need to find the electron's speed if the fields are perpendicular to each other. The magnetic force is balanced by the electric force such that,

$qE=qvB\\\\v=\dfrac{E}{B}\\\\v=\dfrac{1.27\times 10^3}{0.49}\\\\v=2591.83\ m/s$

So, the speed of the electron is 2591.83 m/s.

8 0

2 years ago