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Tasya [4]
3 years ago
13

A small child has a wagon with a mass of 10 kilograms. The child pulls on the wagon with a force of 2 Newton’s. What is the acce

leration of the wagon?
Physics
1 answer:
Zina [86]3 years ago
7 0

PHYSICS

Mass = 10 kg

Force = 2 N

Acceleration = ____?

Answers :

f \:  = m \times a

2 = 10 × a

2 / 10 = a

0,2 m/s² = a ✅

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SP1b.
nata0808 [166]

Answer:

2 m/s^2, west

Explanation:

Vf=final velcoity

Vi=initial velocity

t=timw

a =  \frac{vf - vi}{t}

=

\frac{15 - 25}{5}

= - 2 m/s^2

The - changes direction and makes it opposite

2 m/s, west

3 0
3 years ago
You have discovered a planet that is one-quarter the radius of Earth (Rp = 1/4R⊕) and one-half as massive (Mp = 1/2M⊕). How does
ExtremeBDS [4]
It's 8 times as much as before.
4 0
2 years ago
A bicyclist of mass 90 kg drives around a circle with a centripetal acceleration
Serggg [28]

given,

mass of bicyclist(m)=90Kg

centripetal acceleration(a)=1.5 m/s2

centripetal force(F)=ma= 90×1.5=145 N

5 0
3 years ago
Read 2 more answers
Given two vectors A=4i^+3j^ and vector<br> B=5i^-2j^.find the magnitude of each vector
Flura [38]

Answer:

<em>Magnitude of A=5</em>

<em>Magnitude of B=5.39</em>

Explanation:

<u>The magnitude of Vectors in Rectangular Form</u>

Given a vector v in its rectangular form:

\mathbf{v}=x\hat i+y\hat j

The magnitude of v is:

\mid\mid\mathbf{v}\mid \mid=\sqrt{x^2+y^2}

We are given the vectors

\mathbf{A}=4\hat i+3\hat j

\mathbf{B}=5\hat i-2\hat j

Their magnitudes are:

\mid\mid\mathbf{A}\mid \mid=\sqrt{4^2+3^2}=\sqrt{16+9}=\sqrt{25}

\mid\mid\mathbf{A}\mid \mid=5

\mid\mid\mathbf{B}\mid \mid=\sqrt{5^2+(-2)^2}=\sqrt{25+4}=\sqrt{29}

\mid\mid\mathbf{B}\mid \mid=\sqrt{29}=5.39

4 0
3 years ago
Problem 8: Consider an experimental setup where charged particles (electrons or protons) are first accelerated by an electric fi
yanalaym [24]

Answer:

10581.59 V

Explanation:

We are given that

Magnetic field=B=0.65 T

Speed of electron=v=6.1\times 10^7m/s

Charge on electron, q=e=1.6\times 10^{-19} C

Mass of electron,m_e=9.1\times 10^{-31} kg

We have to find the potential difference in volts required in the first part of the experiment to accelerate electrons.

V=\frac{v^2m_e}{2e}

Where V=Potential difference

m_e=Mass of electron

v=Velocity of electron

Using the formula

V=\frac{(6.1\times 10^7)^2\times 9.1\times 10^{-31}}{2\times 1.6\times 10^{-19}}

V=10581.59 V

Hence, the potential difference=10581.59 V

8 0
3 years ago
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