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Rina8888 [55]
2 years ago
6

A student rides her bike to school. Her school is 5 miles from home. She travels at an average rate of 15 miles per hour. How mu

ch time does she need?
Physics
1 answer:
Elden [556K]2 years ago
3 0

Answer:

.3 repeating hours

Explanation:

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Shawn uses 45 N of force to stop the cart 27 meter from running his foot over. How much work does he do?
Ede4ka [16]

Answer:

W=f×d

w=45×27

w=1215 j.............

3 0
3 years ago
You push a 15N cat in a box horizontally across the floor for 5m. What is the work done on the cat? W=Fx
scZoUnD [109]

Answer:

w=fx

w=(15)(5)

w=75

the angle between the deplacement and N is 0

3 0
2 years ago
An open rectangular tank 1 m wide and 2 m long contains gasoline to a depth of 1 m. If the height of the tank sides is 1.5 m, wh
lara [203]

Answer:

a_y = 4.9\ m/s^2

Explanation:

Given,

Width of rectangular tank, b = 1 m

Length of the tank, l = 2 m

height of the tank, d = 1.5 m

Depth of gasoline on the tank, h = 1 m

\dfrac{dz}{dy}=-\dfrac{1.5-1}{1}

\dfrac{dz}{dy}=-0.5

The differential form with the acceleration

\dfrac{dz}{dy}=\dfrac{-a_y}{a_z + g}

-0.5=-\dfrac{a_y}{a_z + g}

acceleration in z-direction = 0 m/s²

g = 9.8 m/s²

a_y is the horizontal acceleration of the gasoline.

0.5=\dfrac{a_y}{0 + 9.8}

a_y = 9.8\times 0.5

a_y = 4.9\ m/s^2

Hence, Horizontal acceleration of the gasoline before gasoline would spill is equal to 4.9 m/s²

3 0
3 years ago
What is the difference between constant and variable force ?
Nana76 [90]
Constant force - stays the same throughout

Variable force - changes throughout
6 0
3 years ago
A positively charged particle Q1 = +15nC is held fixed at the origin. A second charge Q2 of mass m = 9.5g is floating a distance
Paladinen [302]

Answer:

Q_2=4.4293\times 10^{22}\ nC

Explanation:

Given

charge on the fixed particle, Q_1=15\times 10^{-9}\ C

mass of the second charge, m_2=9.5\times 10^{-3}\ kg

distance between the fixed charge and the floating charge on the top, d=0.25\ m

  • According to the question the second charge is floating just above the fixed positive charge despite of gravity this means that the floating charge is also positive in nature and hence feels the repulsion from the fixed charge which is equal in magnitude to the gravitational force on the charge.

m_2.g=\frac{1}{4\pi\epsilon_0} \times \frac{Q_1.Q_2}{d^2}

where:

\epsilon_0= permittivity of free space

g = acceleration due to gravity

9.5\times 10^{-3}\times 9.8=8.85\times 10^{-9}\times \frac{15\times 10^{-9}\times Q_2}{0.25^2}

Q_2=4.4293\times 10^{22}\ nC

7 0
4 years ago
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