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3 years ago

How does collision time affect impact forces

Physics

1 answer:

Licemer1 [7]3 years ago

5 0

Answer:

A large increase in collision time can mean decrease in impact force. This is essentially what an air bag in your car does. It increases the time over which it stops you and decreases the force. ... In this case, the piano would have a smaller change in momentum and a smaller impact force.

Explanation:

You might be interested in

If a golf ball and a bowling ball are moving and both have the same genetic energy if they move at the same speed which one has

Anastaziya [24]

If they both have the same kinetic energy then they both have the same kinetic energy. otherwise it would be the bowling ball due to factors like weight

5 0

3 years ago

A 21.6−g sample of an alloy at 93.00°C is placed into 50.0 g of water at 22.00°C in an insulated coffee-cup calorimeter with a h

IrinaK [193]

Answer : The specific heat capacity of the alloy $1.422J/g^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$C_1$ = specific heat of alloy = ?

$C_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of alloy = 21.6 g

$m_2$ = mass of water = 50.0 g

$T_f$ = final temperature of system = $31.10^oC$

$T_1$ = initial temperature of alloy = $93.00^oC$

$T_2$ = initial temperature of water = $22.00^oC$

Now put all the given values in the above formula, we get

$21.6g\times c_1\times (31.10-93.00)^oC=-50.0g\times 4.18J/g^oC\times (31.10-22.00)^oC$

$c_1=1.422J/g^oC$

Therefore, the specific heat capacity of the alloy $1.422J/g^oC$

6 0

3 years ago

Kyle is flying a helicopter at 125 m/s on a heading of 325 o . If a wind is blowing at 25 m/s toward a direction of 240.0 o , wh

frosja888 [35]

Answer:

The resultant velocity of the helicopter is $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$ .

Explanation:

Physically speaking, the resulting velocity of the helicopter ( $\vec v_{H}$ ), measured in meters per second, is equal to the absolute velocity of the wind ( $\vec v_{W}$ ), measured in meters per second, plus the velocity of the helicopter relative to wind ( $\vec v_{H/W}$ ), also call velocity at still air, measured in meters per second. That is:

$\vec v_{H} = \vec v_{W}+\vec v_{H/W}$ (1)

In addition, vectors in rectangular form are defined by the following expression:

$\vec v = \|\vec v\| \cdot (\cos \alpha, \sin \alpha)$ (2)

Where:

$\|\vec v\|$ - Magnitude, measured in meters per second.

$\alpha$ - Direction angle, measured in sexagesimal degrees.

Then, (1) is expanded by applying (2):

$\vec v_{H} = \|\vec v_{W}\| \cdot (\cos \alpha_{W},\sin \alpha_{W}) +\|\vec v_{H/W}\| \cdot (\cos \alpha_{H/W},\sin \alpha_{H/W})$ (3)

$\vec v_{H} = \left(\|\vec v_{W}\|\cdot \cos \alpha_{W}+\|\vec v_{H/W}\|\cdot \cos \alpha_{H/W}, \|\vec v_{W}\|\cdot \sin \alpha_{W}+\|\vec v_{H/W}\|\cdot \sin \alpha_{H/W} \right)$

If we know that $\|\vec v_{W}\| = 25\,\frac{m}{s}$ , $\|\vec v_{H/W}\| = 125\,\frac{m}{s}$ , $\alpha_{W} = 240^{\circ}$ and $\alpha_{H/W} = 325^{\circ}$ , then the resulting velocity of the helicopter is:

$\vec v_{H} = \left(\left(25\,\frac{m}{s} \right)\cdot \cos 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \cos 325^{\circ}, \left(25\,\frac{m}{s} \right)\cdot \sin 240^{\circ}+\left(125\,\frac{m}{s} \right)\cdot \sin 325^{\circ}\right)$ $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$

The resultant velocity of the helicopter is $\vec v_{H} = \left(89.894\,\frac{m}{s}, -93.348\,\frac{m}{s}\right)$ .

8 0

3 years ago

Roller coasters accelerates from initial speed of 6.0 M/S2 final speed of 70 M/S over four seconds. What is the acceleration?

choli [55]

Answer:

$a=16\ m/s^2$

Explanation:

<u>Motion With Constant Acceleration </u>

It's a type of motion in which the velocity of an object changes uniformly in time.

The formula to calculate the change of velocities is:

$v_f=v_o+at$

Where:

a = acceleration

vo = initial speed

vf = final speed

t = time

The roller coaster moves from vo=6 m/s to vf=70 m/s in t=4 seconds. To calculate the acceleration, solve for a:

$\displaystyle a=\frac{v_f-v_o}{t}$

$\displaystyle a=\frac{70-6}{4}=\frac{64}{4}$

$\boxed{a=16\ m/s^2}$

3 0

3 years ago

How many milliliters of water at 25.0°C with a density of 0.997 g/mL must be mixed with 162 mL of coffee at 94.6°C so that the r

Andre45 [30]

Answer:

The volume of water is 139 mL.

Explanation:

Due to the Law of conservation of energy, the heat lost by coffee is equal to the heat gained by the water, that is, the sum of heats is equal to zero.

$Q_{coffee} + Q_{water} = 0\\Q_{coffee} = - Q_{water}$

The heat (Q) can be calculated using the following expression:

$Q=c \times m \times \Delta T$

where,

c is the specific heat of each substance

m is the mass of each substance

ΔT is the difference in temperature for each substance

The mass of coffee is:

$162mL.\frac{0.997g}{mL} = 162g$

Then,

$Q_{coffee} = - Q_{water}\\c_{c} \times m_{c} \times \Delta T_{c} = -c_{w} \times m_{w} \times \Delta T_{w}\\m_{c} \times \Delta T_{c} = -m_{w} \times \Delta T_{w}\\m_{w} = \frac{m_{c} \times \Delta T_{c}}{-\Delta T_{w}} \\m_{w}=\frac{162g \times (62.5 \°C - 94.6 \°C ) }{-(62.5 \°C - 25.0 \°C)} \\m_{w} = 139 g$

The volume of water is:

$139g.\frac{1mL}{0.997g} =139mL$

7 0

3 years ago