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VMariaS [17]
2 years ago
9

The largest salami in the world, made in Norway, was more than 20 m long. If a hungry mouse ran around the salami’s circumferenc

e with a tangential speed of 0.17 m/s, the centripetal acceleration of the mouse was 0.29 m/s2 . What was the radius of the salami?
Physics
1 answer:
Musya8 [376]2 years ago
8 0

Answer: 0.09 m

Explanation:

Centripetal acceleration in terms of tangential speed is:

a_c=\frac{v^2_t}{r}

where r is the radius.

It is given that,

centripetal acceleration of the mouse, a_c=0.29 m/s^2

tangential speed , v_t=0.17m/s

Radius of salami is:

r =\frac{(0.17m/s)^2}{0.29 m/s^2}=0.09m

Thus, the radius of the salami is 0.09 m.

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An object moving at a velocity of 32m/s slows to a stop in 4 seconds. What was its acceleration?
Romashka [77]

Answer:

8m/s

Explanation:

a=d/t

a=32/4

a=8 m/s

6 0
2 years ago
Read 2 more answers
Write the equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and
yKpoI14uk [10]

Answer:

The newton’s second law is F=ma

The Gravitational force is F=\dfrac{Gm_{1}m_{2}}{r^2}

Explanation:

Given that,

The equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and Gravitation.

We know that,

Velocity :

The velocity is equal to the rate of position of the object.

v=\dfrac{dx}{dt}....(I)

Acceleration :

The acceleration is equal to the rate of velocity of the object.

a=\dfrac{dv}{dt}....(II)

Newton’s second Laws

The force is equal to the change in momentum.

In mathematically,

F=\dfrac{d(p)}{dt}

Put the value of p

F=\dfrac{d(mv)}{dt}

F=m\dfrac{dv}{dt}

Put the value from equation (II)

F=ma

This is newton’s second laws.

Gravitational force :

The force is equal to the product of mass of objects and divided by square of distance.

In mathematically,

F=\dfrac{Gm_{1}m_{2}}{r^2}

Where, m₁₂ = mass of first object

m= mass of second object

r = distance between both objects

Hence, The newton’s second law is F=ma

The Gravitational force is F=\dfrac{Gm_{1}m_{2}}{r^2}

3 0
2 years ago
Jade and her roommate Jari commute to work each morning, traveling west on I-10. One morning Jade left for work at 6:45 A.M., bu
Veronika [31]

Answer:

Jari

Explanation:

The question requires to know who is traveling faster. This is done by comparing the gradients. The steeper the slope (high gradient), the faster the speed and vice versa.

From Jari's line, the starting point is (0, 0) and another point is (6, 7)

The gradient being change in y to change in x

Change in y=7-0=7

Change in x=6-0=6

Slope is 7/6

For Jade, first point is (0, 10) then another point is (6, 16)

Change in y=16-10=6

Change in x=6-0=6

Slope is 6/6=1

Clearly, 7/6 is greater than 6/6 or 1 hence Jari is faster than Jade

3 0
2 years ago
A huge (essentially infinite) horizontal nonconducting sheet 10.0 cm thick has charge uniformly spread over both faces. The uppe
Nonamiya [84]

Answer:

6.78 X 10³ N/C

Explanation:

Electric field near a charged infinite plate

=  surface charge density / 2ε₀

Field will be perpendicular to the surface of the plate for both the charge density and direction of field will be same so they will add up.

Field due to charge density of +95.0 nC/m2

E₁ = 95 x 10⁻⁹ / 2 ε₀

Field due to charge density of -25.0 nC/m2

E₂ = 25 x 10⁻⁹ /  2ε₀

Total field

E = E₁ + E₂

= 95 x 10⁻⁹ / 2 ε₀ + 25 x 10⁻⁹ /  2ε₀

= 6.78 X 10³ N/C

4 0
2 years ago
The Bernoulli equation is valid for steady, inviscid, incompressible flows with a constant acceleration of gravity. Consider flo
irina1246 [14]

Answer:

p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant

Explanation:

first write the newtons second law:

F_{s}=δma_{s}

Applying bernoulli,s equation as follows:

∑δp+\frac{1}{2} ρδV^{2} +δγz=0\\

Where, δp is the pressure change across the streamline and V is the fluid particle velocity

substitute ρg for {tex]γ[/tex] and g_{0}-cz for g

dp+d(\frac{1}{2}V^{2}+ρ(g_{0}-cz)dz=0

integrating the above equation using limits 1 and 2.

\int\limits^2_1  \, dp +\int\limits^2_1 {(\frac{1}{2}ρV^{2} )} \, +ρ \int\limits^2_1 {(g_{0}-cz )} \,dz=0\\p_{1}^{2}+\frac{1}{2}ρ(V^{2})_{1}^{2}+ρg_{0}z_{1}^{2}-ρc(\frac{z^{2}}{2})_{1}^{2}=0\\p_{2}-p_{1}+\frac{1}{2}ρ(V^{2}_{2}-V^{2}_{1})+ρg_{0}(z_{2}-z_{1})-\frac{1}{2}ρc(z^{2}_{2}-z^{2}_{1})=0\\p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant

there the bernoulli equation for this flow is p+\frac{1}{2}ρV^{2}+ρg_{0}z-\frac{1}{2}ρcz^{2}=constant

note: ρ=density(ρ) in some parts and change(δ) in other parts of this equation. it just doesn't show up as that in formular

4 0
2 years ago
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