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valina [46]
3 years ago
9

a spring has constant 250 Newton per meter. if it is stretched a distance of 0.1m, what is the reatoring force?what if it is str

etched a distance of 5 m? why would this probably not really be possible?
Physics
1 answer:
svetlana [45]3 years ago
4 0
At 0.1m, the restoring force would be 25 Newtons. It would probably not be possible to stretch to 5 meters because that would cause a restoring force of 1250 Newtons.
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A force of 15 newtons is used to push a box along the floor a distance of 3 meters. How much was done?
bezimeni [28]

Answer:

45 J

Explanation:

The equation for work is:

Work=Force*Distance

We can substitute the given values into the equation:

Work=15N*3m\\Work=45Nm\\Work=45J

7 0
2 years ago
Read 2 more answers
A horizontal desk surface measures 1.7 m by 1.0 m. If the Earth's magnetic field has magnitude 0.42 mT and is directed 68° below
AlexFokin [52]

Answer:

The magnetic flux through the desk surface is 6.6\times10^{-4}\ T-m^2.

Explanation:

Given that,

Magnetic field B = 0.42 T

Angle =68°

We need to calculate the magnetic flux

\phi=BA\costheta

Where, B = magnetic field

A = area

Put the value into the formula

\phi=0.42\times10^{-3}\times1.7\times1.0\cos22^{\circ}

\phi=0.42\times10^{-3}\times1.7\times1.0\times0.927

\phi=6.6\times10^{-4}\ T-m^2

Hence, The magnetic flux through the desk surface is 6.6\times10^{-4}\ T-m^2.

3 0
3 years ago
Calculate the force of gravity on the 0.60- kg mass if it were 1.3×107 m above Earth's surface (that is, if it were three Earth
nignag [31]
The gravitational force between two objects is given by:
F=G \frac{m_1 m_2}{r^2}
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is their separation

In this problem, the first object has a mass of m_1=0.60 kg, while the second "object" is the Earth, with mass m_2=5.97 \cdot 10^{24}kg. The distance of the object from the Earth's center is r=1.3 \cdot 10^7 m; if we substitute these numbers into the equation, we find the force of gravity exerted by the Earth on the mass of 0.60 kg:
F=G \frac{m_1m_2}{r^2}=(6.67\cdot 10^{-11}) \frac{(0.60 kg)(5.97 \cdot 10^{24} kg)}{(1.3 \cdot 10^7 m)^2}=  1.41 N
5 0
3 years ago
6. And what's up with ChiN?
n200080 [17]
I don’t know what’s up with chin man ??
4 0
2 years ago
Find the average force exerted by the bat on the ball if the two are in contact for 0.00129 s. Answer in units of N.
Semmy [17]

Answer:

Explanation:

Given

time of contact between bat and ball is t=0.00129 s

suppose u is the incoming velocity and v is the final velocity after collision

m=mass\ of\ ball

Impulse exerted is given by change in momentum of the particle.

Initial momentum P_i=m\times u

Final momentum P_f=m\times v

Change in momentum \Delta P=P_i-P_f

Impulse    J=F_{avg}\cdot t=\Delta P

J=F_{avg}\cdot t=m(u-v)

F_{avg}=\frac{m(v-u)}{t}

F_{avg}=775.2\times m(v-u)\ N                

5 0
4 years ago
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