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3 years ago

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How many ft/s is 60 mph/hr

1 answer:

olga nikolaevna [1]3 years ago

3 0

Answer:

1 Mile per hour = 1.60934 kilometers per hour = 0.44704 meters per second (SI base unit). 1 mph = 0.447 04 m/s. 1 Foot per Second: Feet per Second.

Explanation:

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Vector A has a magnitude of 63 units and points west, while vector B has the same magnitude and points due south. Find the magni

ozzi

Given :

Vector A has a magnitude of 63 units and points west, while vector B has the same magnitude and points due south.

To Find :

The magnitude and direction of

a) A + B .

b) A - B.

Solution :

Let , direction in north is given by +j and east is given by +i .

So , $A=-63i$ and $B=63j$

Now , A + B is given by :

$A+B=-63i+63j$

$| A+B | = 63\sqrt{2}$

Direction of A+B is 45° north of west .

Also , for A-B :

$A-B=-63i-63j$

$|A-B|=63\sqrt{2}$

Direction of A-B is 45° south of west .

( When two vector of same magnitude which are perpendicular to each other are added or subtracted the resultant is always 45° from each of them)

Hence , this is the required solution .

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2 years ago

A glider with mass 0.24 kg sits on a frictionless horizontal air track, connected to a spring of negligible mass with force cons

shepuryov [24]

Answer:

$v=2.556m/s$

Explanation:

From the conservation of mechanical energy

$K_{E1}+U_1=K_{E2}+U_2$

$\frac{1}{2}m*v_1^2+\frac{1}{2}*K*x_1^2=\frac{1}{2}m*v_2^2+\frac{1}{2}*K*x_2^2$

$x_2=0.08m$

$v_1=0 m/s$

Solve to velocity v2

$m*v_2^2=k*x_1^2-k*x_2^2$

$v^2=\frac{k}{m}*(x_1^2-x_2^2)$

$v^2=\frac{5.5N/m}{0.24kg}*(0.54m^2-0.080^2)$

$v=\sqrt{6.54m^2/s^2}=2.556m/s$

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3 years ago

a chamber with a fixed volume of 1.0 meters cubed contains a monatomic gas at 3.00 *10^K. The chamber is heated to a temperature

igomit [66]

Answer:

Explanation:

Given

Volume of fixed chamber $V=1 m^3$

Initial Temperature $T_1=300 K$

Final Temperature $T_2=400 K$

Heat Supplied $Q=10 J$

From First law of thermodynamics

Change in internal energy of the system is equal to heat added minus work done by the system

$\Delta U=Q-W$

as the volume is fixed therefore work

$W=\int PdV=0$

thus $\Delta U=mc_v\Delta T=Q$

$c_v$ for mono-atomic gas is $12.471 J/K-mol$

$n\times 12.471\times (400-300)=10$

$n=0.008018 mol$

and 1 mole contains $6.022\times 10^{23} molecules$

thus No of molecules $=0.008018\times 6.022\times 10^{23}$

No of molecules $=4.82\times 10^{21} molecules$

3 0

3 years ago

All of the following are categories of individual activities except?

aliina [53]

B. Hand-eye Coordination because it can be used in multiple activities

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2 years ago

If the rectangular barge is 3.0 m by 20.0 m and sits 0.70 m deep in the harbor, how deep will it sit in the river?

leva [86]

The harbour contains salt water while the river contains fresh water. So assuming that the densities of fresh water and salt water are:

density (salt water) = 1029 kg / m^3

density (fresh water) = 1000 kg / m^3

The amount of water (in mass) displaced by the barge should be equal in two waters.

mass displaced (salt water) = mass displaced (fresh water)

Since mass is also the product of density and volume, therefore:

<span>[density * volume]_salt water = [density * volume]_fresh water ---> 1</span>

First we calculate the amount of volume displaced in the harbour (salt water):

V = 3.0 m * 20.0 m * 0.70 m

V = 42 m^3 of salt water

Plugging in the values into equation 1:

1029 kg / m^3 * 42 m^3 = 1000 kg/m^3 * Volume fresh water

Volume fresh water displaced = 43.218 m^3

Therefore the depth of the barge in the river is:

43.218 m^3 = 3.0 m * 20.0 m * h

<span>h = 0.72 m (ANSWER)</span>

8 0

3 years ago