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CaHeK987 [17]
3 years ago
5

A 0.111 kg hockey puck moving at 55 m/s is caught by a 80. kg goalie at rest. with what speed does the goalie slide on the frict

ionless ice?​
Physics
1 answer:
madreJ [45]3 years ago
8 0

Answer:

0.076 m/s

Explanation:

Momentum is conserved:

m v = (m + M) V

(0.111 kg) (55 m/s) = (0.111 kg + 80. kg) V

V = 0.076 m/s

After catching the puck, the goalie slides at 0.076 m/s.

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True or False: Inertia is the property that every material object has that causes objects to resist changes in its state of moti
goldfiish [28.3K]
That is True because if it resists it means to not do it and if it is motion, that means that it is resisting movement and that is what inertia is.

The scientific definition for inertia is: "a tendency to do nothing or to remain unchanged"
Glad to help. :)
6 0
3 years ago
How much heat transfer (in kcal) is required to raise the temperature of a 0.850 kg aluminum pot containing 2.00 kg of water fro
hram777 [196]

To solve this problem we will apply the concept related to the heat transferred to a body to reach a certain temperature. This concept is shaped by the energy ratio of a body which is the product of the mass, its specific heat and the change in temperature. For the specific case, it will be the sum of the heat transferred to the Water, the Aluminum and the loss due to latency due to vaporization in the water. That is to say,

\Delta Q = m_{Al} C_p \Delta T +m_wC_w \Delta T  +m_w L_v

Here,

m_{Al}= Mass of Aluminum

C_p= Specific Heat of Aluminum

C_w= Specific Heat of Water

m_w = Mass of water

L_v =Latent of Vaporization

Replacing,

\Delta Q = (0.85)(900)(100-45)+(2)(2000)(100-45)+(0.7)(2258000)

Converting,

\Delta Q = 1842675J (\frac{0.000239006kCal}{1J})

\Delta Q = 440.409kCal

Therefore is required 440.409kCal

8 0
3 years ago
For a vehicle to negotiate a banked curve in poor weather conditions, where the force of friction f = 0, for a given velocity v
djverab [1.8K]

Answer:

R=240m

Explanation:

From the question we are told that:

Velocity v=25m/s

Force of friction f = 0

Angle \theta=15

Generally the equation for  Radius of curvature is mathematically given by

R=frac{v^2}{tan\theta *g}

R=frac{25^2}{tan 15 *9.81}

R=240m

7 0
3 years ago
How do the forces in nuclear and chemical reactions differ?
kvasek [131]

B. Chemical reactions must overcome the strong nuclear force

8 0
3 years ago
Read 2 more answers
A sled is accelerating down a hill at a rate of 1 m s2 . If the mass of the sled is suddenly cut in half and the net force on th
mihalych1998 [28]
We have that F=ma from the 2nd Newton law where F is the force, m is the mass and a is the acceleration. Suppose we have that F' is the new force and m' is the new mass. Then, we have that a'=F'/m' still, by rearranging Newton's law. We are given that F'=2F and m'=m/2. Hence,
a'= \frac{2F}{ \frac{m}{2} } = \frac{4F}{m} = 4\frac{F}{m}
But now, we have from F=ma, that a=F/m and we are given that a=1m/s^2.
We can substitute thus, a'=4a=4*1m/s^2=4m/s^2.
4 0
3 years ago
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